
Copyright N° ii^. 



COPYRIGHT DEPOSIT. 



/ 



DIFFERENTIAL AND INTEGRAL 

<"f CALCULUS J?f 

■" £ 3 & 

WfTH APPLICATIONS 



FOR COLLEGES, UNIVERSITIES, AND 
TECHNICAL SCHOOLS 



BY 



E. W. NICHOLS 



Professor of Mathematics in the Virginia Military Institute, and 
Author of Nichols's Analytic Geometry 



J 



BOSTON, U.S.A. 

D. C. HEATH & CO., PUBLISHERS 

1900 



> 



93037 



Library of Cbngreaa 

Two Copies Received 
DEC 24 1900 

C\ Copyright «tnr 

Ho QJ3 /.//?..... 

SECOND COPY 

Odiwrad to 

ORQEfi DIVISION 

JAN 10 1901 



&<•, 



Copyright, 1900, 
By D. C. Heath & Co. 



PREFACE. 



This text-book is based upon the methods of " limits " and 
"rates," and is limited in its scope to the requirements in the 
undergraduate courses of our best universities, colleges, and 
technical schools. In its preparation the author has embodied 
the results of twenty years' experience in the class-room, ten of 
which have been devoted to applied mathematics and ten to 
pure mathematics. 

It has been his aim to prepare a teachable work for beginners, 
removing as far as the nature of the subject would admit all 
obscurities and mysteries, and endeavoring by the introduction 
of a great variety of practical exercises to stimulate the student's 
interest and appetite. 

Among the more marked peculiarities of the work the follow- 
ing ma}^ be enumerated : — 

i. A large amount of explanation. 

2. Clear and simple demonstrations of principles. 

3. Geometric, mechanical, and electrical applications. 

4. Historical notes at the heads of chapters giving a brief 
account of the discovery and development of the subject of 
which it treats. 

5. Footnotes calling attention to topics of special historic 
interest. 

iii 



IV 



Preface 



6. A chapter on Differential Equations for students in 
mathematical physics and for the benefit of those desiring 
an elementary knowledge of this interesting extension of the 
calculus. 

7. An arrangement of topics admitting of extensive elimina- 
tions without destroying the continuity of the subject. 

8. A clear, open page. 

The author desires to express here his acknowledgments to 

the friends who have aided him in his work. To Chas. M. 

Snelling, A.M., University of Georgia, and to T. H. Taliaferro, 

Ph.D., State College of Pennsylvania, the author's obligations 

are peculiarly great. Not only have they given valuable 

counsel, but they have been largely instrumental in freeing the 

work from typographical errors. 

E. W. NICHOLS. 
Lexington, Va. 



CONTENTS. 



IO-II 
12 

13-14 
15-16 
I7-l8 

19-20 



PART I. DIFFERENTIAL CALCULUS. 

CHAPTER I. 

QUANTITIES. FUNCTIONS. 

ARTS. PAGES 

1-2. Quantity. Classes of 3 

3-4. Constants. Variables ... - 3 

5. Illustrations 3-4 

6—8. Functions. Classes of 4-7 

9. Notation. Examples 7-9 

CHAPTER II. 
FUNDAMENTAL PRINCIPLES. 



Increment. Uniform and Varied Change 10 

Uniform and Varied Motion . 11 

Differential. Illustrations 11— 12 

Rate. Relation between a Differential and a Rate . . . 12-13 

Velocity. Component Velocities 14 

dy 

Significations of -~- ' Remark IS -1 7 

& dx 



CHAPTER III. 

DIFFERENTIATION. 

21-22. History. Differential Calculus. Differentiation .... 18 

23-29. Differentiation of Algebraic Functions. Examples . . . 19-28 

30-32. Differentiation of the Logarithmic Functions 29— 31 

33-34. Differentiation of the Exponential Functions. Examples . 31-35 



vi Contents 



ARTS. PAGES 

35-43. Differentiation of the Trigonometric Functions. Examples 35-40 
44-52. Differentiation of the Circular Functions. Examples . . 40-46 



CHAPTER IV. 
LIMITS. 

53-57. History. Limit. Principles 47~49 

58—59. First Derivative. Examples 49~5 2 

60-67. Differentiation by Method of Limits 52—57 

68. First Derivative as a Fraction 57 

CHAPTER V. 
ANALYTICAL APPLICATIONS. 

Analytical Applications 59—71 

CHAPTER VI. 
GEOMETRIC APPLICATIONS. 

Cartesian Curves. 

69-70. Tangent. Normal. Examples 72—75 

71-72. Subtangent. Subnormal. Examples 75—78 

73-75. Asymptotes. Examples 78-84 

Polar Curves. 

76-77. Tangent. Subtangent 84-86 

78. Normal. Subnormal. Examples 86-88 

79. Asymptotes. Examples 88-90 

CHAPTER VII. 
SUCCESSIVE DIFFERENTIATION. 

80-81. Successive Differentials and Derivatives. Examples . . 91-94 

82. Applications 94~9^ 

83. Leibnitz's Theorem. Examples 96-99 

84. Non-equicrescent Variables. Examples 99-104 



Contents 



vn 



ARTS. 
85-91 

92-93 

94 
95-96 

97 
99-102 



CHAPTER VIII. 
SERIES. 

PAGES 

History. Varieties of. Methods of Development . 105-108 

Maclaurin's Theorem. Examples 108-113 

Euler's Exponential Values of Sine and Cosine ... 113 

Taylor's Theorem. Examples 11 3-1 17 

Bernoulli's Series 117 

Lagrange's Theorem. Tests for Development . . . 11 8-1 24 



CHAPTER IX. 
ILLUSORY FORMS. 

103-105. History. Forms — > — , -• Examples 125-129 

a 00 00 
106-107. Forms — » — • > — . Examples I'so-m 

108. Forms o. 00, 00 — 00. Examples 133-134 

109-110. Forms o", oo°, i°°, o°°, co°°, etc. Examples .... 134-136 



CHAPTER X. 
MAXIMA AND MINIMA. 



111-113. History. Conditions for. Illustration 137-140 

114-116. Methods of Investigation. Suggestions. Examples.. 140-149 
117. Formulae. Problems 149-156 



CHAPTER XL 

PARTIAL AND TOTAL DIFFERENTIATION. 

118-119. Partial Differentials and Derivatives. Examples . . 157-159 

120. Euler's Theorem. Examples 159-160 

1 21-122. Total Differentials and Derivatives. Examples . . 160-166 

123-124. Successive Partial Differentiation 166-169 

125. Successive Total Differentiation 170 



viii Contents 

CHAPTER XII. 
DIRECTION OF CURVATURE. POINTS OF INFLEXION. 

Cartesian Curves. 

ARTS. PAGES 

126-127. Investigation for Direction of Curvature 171-172 

128. Point of Inflexion. Examples . . 172-174 

Polar Curves. 

129-130. Investigation for Direction of Curvature . . ... 174-176 
131. Point of Inflexion. Examples . 176-177 

CHAPTER XIII. 

CURVATURE. EVOLUTE AND INVOLUTE. 

132-133. History. Measure. Circle and Radius of Curvature 178-181 
134-135. Expressions for Radius. Maximum Curvature. Ex- 
amples . . 181-185 

136-140. Evolute. Involute. Examples 185-191 

CHAPTER XIV. 

CONTACT OF CURVES. ENVELOPES. 

1 41 -1 44. Orders of Contact. Examples 192-198 

145-146. Families of Curves. Envelope 198-199 

147-148. Equation of Envelope. Examples ., 199-204 

CHAPTER XV. 

SINGULAR POINTS. 

149-152. Multiple Points. Isolated Points. Point d' Arret . . 205-207 
153. Methods of Investigation. Examples 207-215 

CHAPTER XVI. 
LOCI. 

154-155. Algebraic Equations. Suggestions. Examples . . 216-221 
156. Polar Equations. Suggestions. Examples . . . . 221-224 



Contents 



IX 



PART II. INTEGRAL CALCULUS. 



ARTS. 

157-159 
l60 
l6l 

l62-l64 
165 



CHAPTER I. 
TYPE FORMS. 

PAGES 

Integrals. Integration. Notation 225-228 

Indefinite Integrals. Constant of Integration . . . 228 

Elementary Principles 229 

Type Formulae. Examples 229-243 

Integration by Parts. Examples 243-246 



CHAPTER II. 
RATIONAL FRACTIONS. 

166-168. Fractional Differential and Cases . 

169. Factors Real and Unequal. Examples 

170. Factors Real and Equal. Examples . . . 

171. Factors Imaginary and Unequal. Examples 

172. Factors Imaginary and Equal. Examples 



247-248 
248-251 

2 5 2 ~ 2 55 
2 55~ 2 59 
259-262 



IRRATIONAL FRACTIONS. 

173-174. Methods of Rationalization 262 

175—177. Monomial, Binomial and Quadratic Surds. Examples 262-268 

178. Methods in Special Cases. Examples 268-272 

CHAPTER III. 

BINOMIAL DIFFERENTIALS. 

179-180. Reduction to Form, x m {a + bx n )Pdx 273-274 

181-183. Rationalization. Examples ......... 274-279 

184-187. Reduction Formulae. Examples 279-288 



CHAPTER IV. 
TRIGONOMETRIC INTEGRALS. 



5-189. Trigonometric Formulae. General Rule 289-290 

190. J tan 7 " xdx. . T cof 1 xdx. Examples 290-292 



X 



Contents 



191 

192 

193-194 

196-198 
199 



PAGES 

/ sec" xdx. . f esc" xdx. Examples ' 292—293 

/ ta.n m x sec 1 xdx. . f cot m x esc 11 xdx. Examples . 294—295 

/ sin m xdx. J cos" 1 xdx. Examples 296—298 

J sin m x cos n xdx. Examples 298—301 

Reduction Formulae 3 OI— 3°5 

Reduction to Algebraic Forms. Examples .... 306-311 



CHAPTER V. 
DEFINITE INTEGRALS. 

200-202. Method of Determining the Constant 312-313 

203. Method of Eliminating the Constant 314 

204 Notation. Examples 3 I 4~3 I S 

Applications 316-318 

CHAPTER VI. 
GEOMETRIC APPLICATIONS. 

205-206. Quadrature. Examples 319-324 

207-208. Rectification. Examples 324—329 

209. Surfaces and Volumes of Revolution. Examples . . 329—333 

210. Surfaces and Volumes in General. Examples . . . 333— 33S 



CHAPTER VII. 
SUCCESSIVE INTEGRATION. 



211. Successive Integration 

212-213. Double and Triple Integration . . . 
214. Definite Double and Triple Integration. 



Examples 



336-337 

337-33 8 
33%~339 



CHAPTER VIII. 
GEOMETRIC APPLICATIONS. 

215. Quadrature by Double Integration. Examples . . 340-341 

216. Surfaces and Volumes by Double and Triple Integra- 

tion. Examples 342—346 



Contents 



XI 



CHAPTER IX. 
DIFFERENTIAL EQUATIONS. 

ARTS. PAGES 

217-218. Definitions. Orders. Degrees 347~348 

219. Form, f (x) f x (y) dx-\-§ (x) fa (y) dy = o. Examples^ 348-350 

220-221. Form, f (x,y) dy + <f> (x,y) dx = 0. Examples. . . 350-353 

222-223. Form, dy + Pydx — Qdx 353 _ 354 

224. Form, dy + Pydx = Qy n dx. Examples 354 - 3S6 

225-226. Exact Differential Equations. Examples 356-358 

227-228. Integrating Factor. Examples 358-362 

229. Equations of First Order and nth Degree. Examples . 362-364 

230. Equations of nth. Order. Examples 364-367 



CHAPTER X. 

MECHANICAL APPLICATIONS. 

231-237. Rectilinear Motion 368-374 

238-240. Curvilinear Motion , 374~379 

241-249. Centers of Gravity 379-384 

250-255. Moments of Inertia 384-386 

256-261. Deflection and Slope of Beams 386-393 

262. Strongest Rectangular Beam 393~394 



PART I. 



DIFFERENTIAL CALCULUS. 



DIFFERENTIAL CALCULUS. 



CHAPTER I. 

QUANTITIES. FUNCTIONS. 

1. Quantity. That which can be increased, diminished, 
measured, or in general, anything to which mathematical pro- 
cesses are applicable is called Quantity. 

Time, space, motion, velocity, force, and mass are examples, 
and with these, as with other quantities, we shall have more or 
less to do in illustrating and applying the principles which are 
to follow. 

2. Classes of Quantity. In the abstract science of the Cal- 
culus, as in Analytic Geometry, quantities are divided into 
two general classes, viz., Constants and Variables. 

3. Constants. A constant is a quantity whose value is fixed. 
Constants are usually represented by the first letters of the 
alphabet, a, b, c, etc., or by numbers. 

4. Variables. A variable is a quantity which is, or is con- 
ceived to be, in an actual state of change. Variables are usu- 
ally represented by the last letters of the alphabet, ?/, z>, w, x, y, 
z, etc. 

5. Illustrations. The usual algebraic expression of the 
law subject to which a point moves in generating a circle is 
(x — df ■+■ ( y — 1>f = **» Ifj then, we consider the generating 



4 Differential Calculus 

point to be actually in motion, it is readily seen that its co-ordi- 
nates, x and y, are in a state of variation, and hence, by defini- 
tion (§ 4), are variables ; while the quantity r f which measures 
the distance of the generating point from the centre (a, b) of 
the circle, is fixed in value, and hence (§ 3) is a constant. 

Again : A train leaves Jersey City for Philadelphia, and after 
the lapse of a certain time attains a uniform speed of 45 miles 
an hour, which it maintains until its destination is reached. 
The distance between the train and Jersey City, being at the in- 
stant under consideration in an actual state of change or varia- 
tion, is a variable, while the speed, or velocity (45 miles per hour), 
is a constant. 

Again : A meteor is falling to the earth : Both the distance 
between these bodies and their mutual attractio7i are variables, 
the latter varying inversely as the square of their distance 
apart. 

Again : The volume of water in a cistern which is being filled 
or emptied by a continuous stream is a variable. 

6. Functions. One variable is said to be a function of 
another when its value depends upon that of the latter. Thus : 
The area of a circle (ira 2 ) is a function of its radius (a) ; the 
area of a square is a function of its side ; the expressions x 2 -{- 1, 
log x, sin x, x 3 — log x 2 -\- tan x, as well as all expressions which 
contain x only, are functions of x. In like manner a variable 
is said to be a function of two or more variables when its value 
depends upon their values. Thus, the area of an ellipse (nrab) 
is a function of its semi-axes (a, b) ; the area of a rectangle is a 
function of its base and altitude; the volume of a rectangular 
parallelopiped is a function of its three dimensions ; the expres- 
sions x 2 -\- y 2, — a 2 , sin x + tan y, x s — log y 2 are functions of x 
and y ; the expressions xy -f- z 5 , z -f- log xy, xyz 2 , etc. are func- 
tions of x, y, and z. 

When a function of one or more variables involves no condi- 



Quantities Functions 5 

tion or conditions, the variables are indepe7ide?it of each other ; 
that is, we may assign to each any value we please. Such is 
the case in all the illustrations given above. If, however, some 
condition is involved, as, for example, the equality to zero of 
any one of these expressions, then one variable at least must be 
dependent for its value upon the values assigned the others. 
Let us illustrate : In the function of x and y, x 2 + y 2 — a 2 , x 
and y are independent variables — no condition being involved. 
But suppose we write x 2 + f — a 2 = o, then the range of values 
which may be assigned to x or y is at once limited, and we 
can no longer assign arbitrarily any values we please to both, 
but must assign values to one only, and ascertain from the 
equation the value of the other. The variable to which values 
are assigned is called the independent variable ; the other, which 
now represents the function, is called the dependent variable. 

7. General Classes of Functions. Functions are either Alge- 
braic or Transcendental : 

I. Algebraic Functions are those which involve only the six 
fundamental operations of algebra, viz. : Addition, Subtraction, 
Multiplication, Division, Involution, and Evolution, with constant 

indices. Thus, x + y, x 2 — yy, 1- V* 3 are algebraic func- 
tions. 

II. Transcendental Functions, embracing all functions other 
than algebraic, are subdivided into various classes, the more 
important being : 

1. Trigonometric Functions, such as sin x, tan x, sec x, etc. ; 

2. Circular, or Inverse Trigonometric Functions, such as sin -1 #, 
tan -1 .*, sec - x x, etc. ; 



3. Logarithmic Functions, such as log x, log (x -f- v'x 2 — y 2 ), 
etc. ; 



6 Differential Calculus 

4. Exponential, or Inverse Logarithmic Functions such as a x , 

x y , (u + v) z . 

8. Special Classes of Functions. Both algebraic and trans- 
cendental functions are further subdivided, the subdivisions 
being dependent upon the standpoint from which they are 
viewed. 

I. Explicit and Implicit Functions. 

1. The Explicit Function, as y = x 3 — ax + b, or y = log x, 
or y = x 2 z + sin v, where the simple fact that the equation is 
solved with respect to one of the variables which enters it, 
indicates explicitly that the first member (y) is a function of the 
variables which enter the second member. 

2. The Implicit Function, as y — x 3 + ax — b = o, or y— log 
x = o, or y — x 2 z — sin v = o, where the fact that one of the 
variables is a function of the other is implied from the condi- 
tion of equality to zero. 

II. Increasing and Decreasing Functions. 

1. The Increasing Function, as y= sx + b (s = positive quan- 
tity), for as x increases, the function^ increases also. 

2. The Decreasing Function, as y = — sx + b (s = positive 
quantity), for as x increases, the function y decreases. 

It should be carefully observed that the terms iiicrease and 
decrease are here used in an algebraic sense. In the common 
parabola y = + *^2px for example, y is an increasing or a de- 
creasing function of x, according as we use the upper or lower 
sign before the radical ; in other words, y is an increasing 
function of x in the first angle, and a decreasing function of x 
in the fourth. 

III. Continuous and Discontinuous Functions. 

1. The Continuous Function of a variable is a quantity that 
changes gradually, and passes through every intermediate value 



Quantities Functions 7 

from an initial to a final value, as the variable that enters it 
passes through every intermediate value from its initial to its 
final value. 

Thus in the equation of the line, y = sx -f- b, y is a continue 
ous function of x\ for as x increases gradually, and passes 
through all intermediate values between — oo and -f- °° , y also 
changes gradually, and passes through all intermediate values 
between — oo and + oo. Again, in the circle y= ± \/a 2 — x 2 , 
y is a continuous function of x between the limits x = ± a. 

2. The Discontinuous Function — as in the hyperbola y = 

,b \ ... 

±- Vx^—a 2 . Here y is a discontinuous function of x between 

the limits x = ±a. For values of x> ± awe, readily see that 
y is a continuous function of x. In fact, functional forms fre- 
quently occur which are continuous between certain limits of 
the variable which enters it, and discontinuous between other 
limits of that variable. The Differential Calculus, however, 
has only to do with variables between their limits of continuity. 

9. Notation.* The equation y =/(x) is a symbolic expres- 
sion of the sentence u y is an explicit function of x." Similarly 
y =f(x, z, v) is to be read " y is an explicit function of x, z, and 
v." Implicit functions are also capable of general representa- 
tion. Thus,y*(x, y) = o means that x and y are implicit func- 
tions of each other. 

If the same functional symbol occurs more than once in the 
same operation it is understood to refer to the same function. 
If the symbols are different, then the functions to which they 
refer are different. Thusy(x) and (f> (x) would, in the same 
operation, indicate that the functions of x to which reference 
was made were different. 

* The notation <f> (_r) to indicate a function of x was introduced by John Bernouilli, the 
elder, in 1718 ; but the general adoption of symbols like_/, F, <$>,$, . . .to represent func- 
tions, was mainly due to Euler and Lagrange. 



8 Differential Calculus 

If a particular value is assigned a variable which enters a 
function, and we wish to express in general notation the re- 
sulting value of the function, we substitute for the variable this 
particular value. Thus, if we wish to indicate what ./"(%) be- 
comes when x is equal to a, or b, or o, we write /(a), or / '(d), 
or f (o), as the case may be. Another method, where the par- 
ticular function with which we have to deal is given, is to place 
the value of the variable as a subscript to a semi-bracket placed 
on the right of the function, and equate this to the result ob- 
tained by substituting the value of the variable in the function. 

Thus 

x — 



X + 2 



x 2 + 2 ax 
= o; 

2 x — a 



3 a. 



EXAMPLES. 

1. If f(x) == x 2 — 5 x -f- 6, show that 

/(l)=2, /(-2)=20, /(I) = 3 t, /(2)=o, 
f(S) = °» f{ X - l) = X 2 — 7 2C -f 12, 

y^2 .%') = 4^ — io x -f- 6. 

2. If /"(#) = (# -f- i) (x — i) (# — 2), show that 

/(- I) =/(l) =/(2), - 3/(3) = 2/(- 2). 

3. If/00 = <? - *" r , show that /( 3 7) = [/(j)] 3 - 3 /(x>, 

4. If <£ (x) = a x , show that [<£ (x)] 2 = <£ (2 #). 

In the following implicit functions make y an explicit function 
of x. 

5. jy 2 — 2 xy + x 2 = o y = x 

( a _j_ ^2 

6. log a y = 2 log a (0 + x) - 1 j> = -^- • 

7. Is y an increasing or decreasing function of x in the 
function given in Ex. 5 ? Is it continuous ? 



Quantities Functions 
8. Show that the following equalities are true : 



9 



X 2 -\- X -\- 1 
2 X — I 



Jl 



3> 



2 sin x cos x 
cos x — sin x 



= o, 



csc'x 



2 
IT 



J2 



9. In the equation x (y — 2) + y — c = o, show that y is 
not a function of x when ^ = 2 . 

10. Show that y is not a function of x in the equation 
sin x sin \ x -f- cos ^ x cos jc 



jr = 



cos -J-.* 



10 Differential Calculus 



CHAPTER II. 

FUNDAMENTAL PRINCIPLES. 

10. Increment. The increment of a variable is the amount of 
its change in any interval of time. 

We can always ascertain this amount by taking the algebraic 
difference between the values of the variable at the beginning 
and at the end of the interval — always subtracting the former 
from the latter. If the increment thus ascertained is positive, 
the variable is increasing ; if negative, it is decreasing. See 
§ 8, II. The increment of a variable is usually denoted by the 
symbol A placed before the variable. Thus Ax means ' the in- 
crement of x,' and is to be so read. 

11. Uniform Change. Varied Change. When a variable so 
changes that its increment is numerically the same in all equal 
intervals of time, its change is said to be uniform. 

In all other cases its change is said to be Varied. 

A dcbaBabcd 

i i i i 1 i i I 

Fig. a. 

Thus let AB represent graphically the state of a variable (u) 
at any instant, and let Ba, ad, be, etc., represent its increments 
in any successive equal intervals of time ; then if 

Ba = ab = bc= etc., 

the variable AB (u) is uniformly changing. Otherwise the 
variable is varied in its change. If u is an increasing variable, 
the increments Ba, ab, be, etc., are positive; if decreasing, the 
increments are negative. § io. 



Fundamental Principles 11 

Again, suppose a bucket in the form of an inverted conical 
frustum is being filled from a hydrant in such a manner that 
the depth of the water increases by one inch in every second — 
and proportionately for any other interval of time — then the 
depth of the water is a variable in a state of uniform change, 
while the volume of the water is a variable in a state of varied 
change. Had we assumed a cylindrical bucket, then the volume 
as well as the depth of the water would have changed uniformly. 

12. Uniform Motion. Varied Motion. A point is said to have 
uniform motion when the dista?ice over which it passes, estimated 
from any point in its path, is a variable in a state of uniform 
change. When this distance is varied in its change, the motion 
of the point is also Varied. 

Thus, Fig. a, let us suppose AB to be the path of a flowing 
point, which at the instant of consideration has reached the 
position B. Then if the distance AB is a uniformly changing 
variable, the motioii of the point B is uniform. Otherwise the 
motion of B is varied. 

Corollary. The direction in which a distance is changing is 
determined at any instant by the direction of motion of the flowing 
point at that instant. 

13. Differential. The differential of a variable is the increment 
it would take on in any interval of time, if its change beca?7ie unL 
form at the beginning, and continued so throughout that interval m 
The differential of a uniformly changing variable is obviously 
the actual increment it takes on in any interval. 

The usual notation for representing the differential of a varia- 
ble is the letter d placed before the variable. Thus du, read 
' differential of «,' indicates the operation of taking the differen- 
tial of the variable u. It should be remembered that the sym- 
bol d before u is not a coefficient, but a symbol of operation, and is 
entirely analogous to the symbols sin, cos~ x , log, in the expres- 
sions sin u, cos -1 u, log u. 




12 Differential Calculus 

Cor. The differential of an increasiiig variable is positive, and 
that of a decreasing variable is negative. (§ 10.) 

14. Illustrations. To illustrate the relation between an in- 

crement and a differential, 
as well as to secure a clear 
conception of each, let y = 
f(x) be the equation of any 
locus such as OPT, Fig. i, 
and let u represent the vari- 
able area bounded by the 
curve, the x — axis and the 
terminal ordinate PM, as 
that ordinate moves uniformly to the right, changing its length in 
obediance to the law expressed in the equation y =f(x). Let 
x = OM = x' , and let MN be the increment of x (Ax) in any 
interval of time, beginning at the instant when x = x r . Then 

Ax = MN, Ay = RT, 
and An = area PMNT. 

Since PM moves uniformly the distance OM{= x) changes 
uniformly ; hence dx = Ax = MN. 

Again, du is by definition the increinent that u (OPM) would 
take on if it became a uniformly changing variable, and so con- 
tinued throughout the interval of time. But this supposition of 
uniform change in // (OPM) obviously requires the ordinate 
PM to remain constant in length throughout the interval. 
Hence 

du = area PMNR. 

But area PMNT- area PMNR = area PRT; 
Au — du = area PRT. 

15. Rate. The Measure of the Rate of Change of a variable 
or, more simply, its Rate, is the increment it would take on in a 



Fundamental Principles 13 

unit of time if its change became uniform at the beginning and con- 
tinued so throughout that unit. 

The rate of a uniformly changing variable is the actual incre- 
ment it takes on in a unit of time. 

Thus when we speak of a body falling at the rate of 50 feet 
per second, we mean that the variable distance through which 
the body has already fallen would take on the increment 50 feet 
in the next second, if its change became uniform at the begin- 
ning of the second, and continued so throughout. 

Or, when we speak of a passing train as moving at the rate 
of 40 miles an hour, we mean that if its distance from some 
point in its path (say the last station) became at the instant of 
speaking a uniformly changing variable, and continued so for 
an hour, that it would take on the increment of 40 miles. 

16. Relation Between a Differential and a Rate. It will be ob- 
served that the only difference between the definition of ' a dif- 
ferential ' (§ 13) and that of ' a rate ' (§ 15), is in the use of the 
term " interval of time " in the former, and the term " unit of 
time" in the latter. If, therefore, the "interval of time" is 
taken as the " unit of time," the rate of a variable and its differ- 
ential are the same. If the " interval of time " is not taken as 
the " unit of time," let dt = that interval (since time (t) changes 
uniformly, dt can represent any interval or increment of time), 
and let du be the corresponding differential of a variable u ; 

then 

du , N 

"=Tt (I) 

where r = rate of u. If dt= unit of time, we have 

r = du 

as explained above, i.e., the rate of a variable is its differential 
for a tmit of iiine. 
Referring to the last illustration of § 15, suppose we say that 



14 



Differential Calculus. 



the train will travel 160 miles in the next 4 hours at its present 
rate of travel, then 

du = 160 miles and dt — 4 hours, 
du 160 m 



and 

as before. 



dt 4 h 



= 40 miles per hour 



17. Velocity. Velocity is the rate of change of a distance. Let 
s = variable distance traversed by a point, and let v = its 
velocity at any instant; then, § 16 (1), 



ds 



(.') 



18. Relation between the velocity of a point in its path and its 
component velocities in the direction of rectangular axes. 

Let y =f(x) be the equation of any locus, as APB, when re- 
ferred to rectangular co- 
ordinates. Let us further 
suppose that the point 
which generates this locus 
is at the instant of con- 
sideration at P, (x, y). 
We wish to compare the 
velocity of P in its path 




Fig. 2. 



with its component velocities in the directions of X and Y. 
In other words, we wish to compare the rates of change of the 
distances AP (= s), NP (= x) and MP (== y). 

The direction of change of the varying distance AP(= s) is 
at the instant of consideration the direction of motion of its gen- 
erating point P, § (12) Cor.; but the direction of motion of 
P is at the instant in the direction of the tangent PT. Hence 
the direction of change of the distance AP is at the instant in 
the direction PT. Now, assuming the distance AP (=•*") to 
become a uniformly changing variable at the instant of reach- 



Fundamental- Principles 15 

ing the value AP, lay off any distance FT, in the direction of its 

change, i.e., a!o?ig the tangent FT, to represent the increment it 

would take on under this supposition in the interval of time dt\ 

then, § 13, we have 

ds = FT 

But if AP (= s) becomes a uniformly changing variable in 
the direction FT, the co-ordinates of P, i.e., the distances NP 
(= x), MP (= y), also become uniformly changing variables, 
and would take on the increments PR and FT, respectively, 
in the same interval of time, dt. Hence 

PF = dx, FT= dy. 

From the right triangle PTR we have 

FT 2 = PP* + FT 2 ; 

i.e., (dsf = (dxf + (dyf ..... (3) 

i.e., § 17, 2, The square of the velocity of a poi?it in its path is 
equal to the sum of the squares of its components in any two rect- 
angular directions. 

Cor. Let TSX = a ; then, since TPF = TSX = a, we have 
from the right triangle TPF the following important relations : 

dx , N 

-Y- = cos a (4) 

as 

^j- = sm a (5) 



-^ = tan a ....... (6) 

ax 



19. Signification of Q 
ax 



16 Differential Calculus 

I. Geometric Signification. Every relation between two vari 
ables which can be expressed in the form of an equation, 
y =/(x), can in general be represented geometrically by a 
plane locus. Hence the ratio of the differentials of these vari- 
ables ( -j- ) ought to admit of geometric interpretation. We 

see from (6) of the preceding article that it does admit of such 
interpretation ; for, Analytic Geometry, p. 2 5 , 

tan a = s = slope of tangent TS. 

But Slope of TS = slope of APB [y =/(#)] at (*> y\ 

dy 
hence generally, — = slope of y =f(x) at (x, y). 

II. Analytical Signification. 

dy . 

since -r = rate of y. 

dt 

dx . 

and -r = rate 01 x, 

dt 



and since 



we have 



dy 

dt dy 

dx dx 
"dt 

dy rate of y 

dx rate of x 



Hence the ratio of the differentials of two variables correspond- 
ing to the same interval of time is equal to the ratio of the rates of 
those variables at the beginning of that interval. 

Cor. If rate of x be taken as unit rate. Then 

-f- = x-rate of y. 
dx 



Fundamental Principles 17 

20. Remark.* The terms "unit of time" and "interval of 
time "' as used in preceding articles do not refer to any specific 
portion of time — their values, whether great or small, not being 
considered. In the abstract science of the Calculus, time is a 
" foreign element " ; but as all change occurs in time, it is essen- 
tial to the comparison of the rates of variables related in any- 
given way that the " unit of time," or " interval of time," used 
for this purpose should be understood to be the same. 

Again, as any "interval of time *' may be taken as a "unit 
of time " it will be found convenient to take 

dt = unit of time. 

Unless otherwise stated we shall so consider it in what ensues. 



* Objection has frequently been made to Newton's method of fluxions, that it introduced 
a foreign idea, namely that of motion into geometry and analysis. This objection was an- 
swered by Newton when he stated that all his method contemplates is that one of the vari- 
ables should increase uniformly (aequabili fluxu) as we conceive time to do. 



*s : 



18 Differential Calculus 



CHAPTER III. 

DIFFERENTIATION. 

History. — It is not certain whether the Calculus was first discovered 
by Sir Isaac Newton (1642-1727) or simultaneously and independently by 
Newton and Gottfried Wilhelm Leibnitz (1646-17 16). The facts — elicited 
after a bitter controversy extending throughout the eighteenth century — 
are briefly these : 

1. Newton communicated his discovery to friends and in manuscript in 
1669, although his method was not published until 1693. 

2. Leibnitz published a memoir on the Calculus in 1684, yet the earliest 
use of its method in his note-books is dated 1675. 

3. In 1849, Gerhard discovered among Leibnitz's papers a manuscript in 
Leibnitz's handwriting of extracts from one of Newton's papers, together 
with notes on their expression in the differential notation. A copy of this 
manuscript, it is known, had been sent to Tschirnhausen in May, 1675 • an< ^ 
as he and Leibnitz were engaged on a piece of work at the time, it is pos- 
sible that these extracts were made then. On the other hand, the extracts 
may have been made from the printed copy in 1704. 

4. It is certain that Leibnitz enjoyed for fifteen years and unchallenged 
the honor of being the inventor of his Calculus. Newton himself rendered 
him that credit in the first two editions of his Principia. 

The problem of the Calculus as stated by Newton was : Given the relation 
of the fluents (= variables) to find the relation of their fluxions (= rates). 
This is equivalent to differentiation. Leibnitz's notation being preferable 
to that of Newton has been generally adopted in treatises on the subject. 

21. The Differential Calculus is the science of rates, and its funda- 
mental object is to determine and compare the rates of related 
variables. 

22. Differentiation is the process of determining a dijferefitial. 



Differentiation 19 



ALGEBRAIC FUNCTIONS. 



23. The differential of a constant is zero. 

Let c be any constant. By definition, § (3), a constant is a 
quantity whose value is fixed ; hence, there can be no incre- 
ment to its value in any unit of time. 

Hence dc = (1) 

24. The differential of the sum of two variables is the sum of 
their differentials* 

To prove 

d(u +7') = du 4 dv, 

in which u and v are the values of two varying quantities at 
the same instant, and du and dv are their differentials. 

Then u 4 v, 

v 4 du + (v 4- <&)» 
2* 4- 2 //# + (# + 2 */z/), 

are the values of the function (u 4 #) at the instant and at the 
ends of two equal units of time following that instant under the 
supposition of uniform change in u and v throughout the two units 
of time. Subtracting the first value from the second, or the 
second value from the third, we find that the difference in either 
case is 

du 4 dv. 

But this difference is the increment taken on by the function 
(u 4- z ; ) in the same unit of time, and since it is the same for 
all equal units of time the function is a uniformly changing 
variable at the instant. Hence, § 13, this increment is its 
differential ; 
i.e., d (u + if) = du 4 dv (2) 

* Argobast (a.d. 1800) was the first to separate the symbol of operation from that of 
quantity. Francois in 1812, and Servois in 1814, were the first to give correct rules on the 
subject of operation. 



20 



Differential Calculus 




Cor. i. Since the differential for a unit of time is the rate 
of the variable, § 16, we may interpret (2) as follows: The rate 
of the sum of two variables is the sum of their rates. 

Cor. 2. The differential of any polynomial is the sum of the 
differentials of its term. Thus 

d (u + v — w -f- z) = du + dv — dw + dz. 

25. The differe?itial of the product of two variables is equal to 
the sum of the products arising from inultiplying each variable by 

the differe?itial of the other. 
To prove 

d (uv) = udv + vdu. 

Let u and v be any two vari- 
ables, and, at the instant of 
cojzsideration, let their values 
be represented by the sides 
AB and AD of the rectangle ABCD. Then at the instant 

uv = area ABCD. 

Let BB' = du, DD' = dv, i.e., let BB' and DD r represent the 
increments that the variables u (AB) and v (AD) would take 
on in a unit of time if their changes became uniform at the 
instant of consideration and so continued for the unit of time. 
Now d (uv) [= ^/(area ABCD)\ is the increment that the rect- 
angle would take on in the same unit of time provided its change 
became wiiform at the instant and continued so throughout the 
unit of time. The change in the rectangle ABCD would obvi- 
ously become uniform if it took on the increment 

area DD'C'C + area BB'C'C. 

Moreover, since we have supposed u and v to take on the in- 
crements du (BB') and dv (DD), the rectangle uv (ABCD) can 
change uniformly in no other way. Hence 



Fig. 3. 



Differentiation 21 

^(area ABCD) = area DD'C'C + area BB'C'C; 

i.e., d(wv) = ud'V + <Wa (3) 

Cor. 1. If in (3) we make z/ = £ = a constant, we have 

d {lie) = z/^/<f + rdfo. 
Hence § 23 (1), d(cu) = cdu (4) 

i.e., The differential of a constant multiplied by a variable is 
equal to the product of the constant and the differential of the 
variable. 

Cor. 2. If in (3) we make v = vw, we have, 

d (uvw) = ud (vw) -f- vwdu 

= u (vdw -\- wdv) + vwdu 

= uvdw + uwdv + vwdu. 
Similarly we may prove 

d(u c v c wz) = wvwdz + u c uzd c w + u c wzd c v + c v c wzdu (5) 

and so for any number of variables. 

Hence, 77z^ differential of the product of any number of vari- 
ables is equal to the sum of the products arising fro?n multiplying 
the differential of each variable by the product of all the others. 

26. The differential of a fraction is equal to the dejiominator 
into the differential of the numerator minus the numerator into 
the differential of the denominator divided by the square of the 
denominator. 

To prove 







it): 


vdu — udv 

1? 


u 

Let - = z, 

V 


then 


u = vz and 








du - 


= vdz -f- zdv. 


Replacing 


z by 


. u 
its value - , 

V 


we have 






du - 


, fu\ u . 
= vd 1 — M — dv. 
\v V 



§ 2 5> (3) 



22 Differential Calculus 

du dv 

I u\ V 
Hence, solving, d f - ) = ; 

1 /u\ <vdu — ud*v ,, x 

•■•nw = — * — (6) 

Cor. i. It v = c = a. constant, formula (6) becomes 

(u\ cdu — tide 

• ,$-*.. .... . . . . (7) 

This is as it should be, since - = -u, which being differentiated 
by formula (4) gives d f- ,u\ = - .du = — . 

Cor. 2. If u = c = a constant, then formula (6) becomes 

*©=-c$ ....... (8) 

27. 7/^ differential of a variable with a constant exponent is 

qual to the product of the exponent, the variable with its exponent 

diminished by o?ie and the differential of the variable. 

To prove 

d (u n ) = nu n ~ 1 du. 

1. Let n be a positive integer. 
Then u n = u. u. u. u. to n factors. 
Hence, 

d (u n ) = d (u. u. u. u to n factors) 

= u n ~ x du + u n ~ 1 du + u n ~ x du to n terms. § 25, (5) 
Therefore 

d (a*) = nu n ~ x du (9) 



Differentiation 23 



m 

2. Let n be a positive fraction and equal to — • 

m 

Let y — up, then 

yP = U m . 

Hence, py p ~ x dy = mu m ~ 1 du. Equa. (9) 



m u 



m — 1 



dy = — r du. 





Substituting lot y its value and reducing, we have 

_ m -_ 1 

d(up) = — up du. 

Hence the rule applies in this case. 

3. Let n be negative and equal to — m. 
Let y = u~ m = — , then, § 26. Cor. 2, 

mu m ~ l du 

*y= ^- s 

i.e., du~ m = — mu~ m ~ x du. 

Hence the rule applies in this case also. 

4. Let n be incommensurable. 

This case cannot be considered here as equa. (9) was de- 
duced under the supposition that n was commensurable. The 
rule holds good in this case also, as will be shown in a subse- 
quent article. See § 32. 

28. Differential Equation. Differential Coefficient. 

Let y = x™, then by differentiation 

dy = mx m ~ x dx ; 



dy 



m — l 



hence, -f- = mx 

dx 

The first of these equations is called The First Differential, or 



24 Differential Calculus 

The First Derived Equation, of the equation y = x m , and the 
second is called the First Differential Coefficient, or First Deri- 
vative, of the same equation. Both of these equations, it will 
be observed, contain a new function of x, viz., mx m ~ 1 . Hence ? 
in general, if y =/(x), 
then dy = f\x) dx, 

and g =/'(*) 

are, respectively, the first differential equation and the first differ- 
ential coefficient of the equation y =/(x). 

Cor. Since dy and dx have the same or different signs, ac- 
cording as y and x are increasing or decreasing functions of 

each other, it follows that — is positive or negative, according 

as j> is an increasing or a decreasing function of x. § 13, Cor. 

29. Formulas. 

dc = (a) 

d(cu) = cda (p) 

d(u + c v) = du + d<v (c) 

d(u e v) = ud e v + 'vda (d) 

d (wvw) = u c vd t w + u e wd e v +■ e v c cvdu (*) 

j (u\ e vda—ad c v , > x 

d U~ — ^r~ (/) 

d(u n ) = nu n - 1 du Cf) 

Note. — These formulas are collected here for ease of reference. They 
should, however, be carefully committed to memory. 

EXAMPLES. 
Differentiate : 

1. 5 * + 6. 

d(£x+ 6) = d( 5 x) + d(6) (c) 

= 5 dx. Ans. (b) 



Differentiation 



25 



2. x* — nx 2 + c. 

d(x* - nx 2 + c) = d (x 4 ) - d (nx 2 ) + </ (c) 
= 4 jc 3 ^x — nd (x 2 ) 
= 4 # V.r — 2 ;z^</x 
= 2 x (2 ^ 2 — ^) dx. Ans 

3. #* (*"• — X s ). 
d[x$ (x m — x)] = tf* ^(> m — x) + (% m — a;) //(**) 

= ^i (mx 7 "" 1 — 1) rt& +(jt w — #) \x~^dx (g), (c) 



(') 



(d) 



= \xh 



(mx m - x — 1) + 



v* 



2 V.* 



?]*. 



(2 m -\- i) x m — 3 x 
2 Vx 
or, d\x\ (x™ — x)] = d(x m + l * — #§) 



dx. Ans. 



3 1 



= (m 4- i) 3? + *"*<&? — -a&/# 



<& (') 



-[ 



2 w 4- 1 



1 3 ^ 



2 2 

(2 #z + 1) x™ — 3 ^ 



^r 



v, 



//x as before. 



2 v.# 



\/.r 2 (jc n + nix?) 



4. 

\/^: 2 (vr" + mx 2 ) 
x$ 



\= d\x* (x* 



4- mx 2 ])] 



6w+l 



= d(x 6 + nix**-) 
6n 4- 1 6 -^±- 1 -1 



^ 



r 3 



<&" + -7- #** l dx (g), (c) 



6w— 5 



_ (6n 4- 1) jt: 6 +13 



mx* 



dx 



dx 
= [(6 n 4- 1) *" 4- 13 w* 2 ] _ . Ans. 

6 V* 5 



26 Differential Calculus 

As no special rule has been deduced for radical quantities, 
they must always be expressed as quantities affected with frac- 
tional exponents. It will be observed also that before differen- 
tiating in the above example, the expression was first simplified. 
This should always be done wherever practicable — not as a 
matter of principle, but with a view of simplifying the process. 

( X n (Z + 2) p 
I 7 

_ fd \x n (z + 2) p I - x n (z + 2)Pd{y>) 

Too \J ) 

_f \x n p (z + 2)P~ 1 dz + (z + 2) p nx n ^ l dx} — x n (z + 2) p $y 4 dy 

y° 

x n ~ l (z + 2) p ~ 1 \pxydz -\-{z-\-2) (nydx — 5 xdy\ 

6. x 2 -\- y 2, = a 2 , and find value of dy. 

d(x? +f) = d(a)\ 

2 xdx + 2 ydy = 0, — - 

x 
,\ dy = </.*. ^4«j. 

7. y 2 = 4 ##, and find value of ^v. 

d(y 2 ) = */(4tfx), 

2 _>v/y == 4 # </x, 

2 # 
.*. dy = — */#. ^4«j. 

y 

8. .-j 2 ^ 2 -f- ^p 2 = # 2 ^ 2 > and find value of dy, 

IPx 

Ans. dy = =- </x. 

9. xy = m, and find value of dy. 

y 

Ans. dy — dx, 

x 



Differentiation 27 

10. y = (m + nx) x s . Ans. dy = (^nx + 3 m) x?dx. 

x™ . , mx m ~ 1 

11. y = 7 r: • ^te. «y = 7 rT^*. 

•^ (1 + x) m J (i + jt:)" i + 1 

This fraction may be placed in the form x m (1 + x)~ m and 
differentiated as a product. So with all fractions. 

J 12. j> = (1 + #) Vi — #. ^4«i". ^V == dx. 

2 Vl — x 

1 -I - # 

This product may be placed in the form — - and dif- 
ferentiated as a fraction. Similarly for all products. 

13. y = xi (x$ + i)i ^?w. dy = 3 — ^r. 

12 V-* 2 Vx2 -f- 1 

14. y = x 5 (m -\- 3 x) s (m — 2 .x) 2 . 

^;/j". */y = 5 -x 4 (#* + 3 #) 2 (m — 2 x) (m 2 + 2 #z.r —12 .x 2 ) ^r. 

* 

x™ mx m ~ x 

15. _v = Ans. dy — -, T- n dx. 

J 1 + xT J (1 + * w ) 2 

r^ F ,, 7 4 V* 3 + 

16. y= \ xr -\- aVx. Ans. dy = — dx. 

4 \fx \ x 2 + # V# 

1 — . rs-5 ^ 7 V0 + 3 ** 7 

2 V* 



18. y = y ^— ^ • ^«j. ^ = 



^r 



+ x (1 + #) Vi -* 2 

1 # — Vi + x 2 , 
^19. 7= • ^4«i-. </y = - — dx. 

x + Vi + x* Vi + * 2 

Vi +*+ Vi -* H-Vi-^ 2 , 
20. y = — — = • ^/zx. dy = — //x. 

Vi + x — yi — x x 2 Vl — jc 2 

Rationalize the denominator before differentiating. 



28 Differential Calculus 



Vi -h x 2 + Vi — x 2 
21. y = — = • 

Vi + x 2 — Vi — x 2 



Ans. dy = § ( i + * ) 

x \ Vi — x*/ 



dx. 



x + Vi -+- -t 2 
f 22. y = =+ 

/• Vi + ^: 2 — ^ 



^4^. //>/ = 2 ( Vi + ^ 2 + x) I + i^^U £ 

\Vi + x 2 I 

Write the first derivatives of the following : 

m dy 6 mx 

23 * J=Z {7i 2 + x*f' dx = ~(V +.X 2 ) 4 ' 

2 V* ., dfy ^ (i — X 2 ) 
24. V = - • ^tfj. -f- = -^ ^r • 

3 + ^ <& (3 + .x 2 ) 2 VS 

v 25. jy = V • Ans. -f= — 



— ' ^f/W. — = — ~ 

I + V* " X 2 (i + V#) \g)(l — X) 



26. j; = 



(x z — 2 .#) ^ m 

X 2 



iyJ[\ 



dy , 
Ans. — — (m -\- 1 ) x m — 2 (m — 1 ) x m ~ 2 . 
dx J ' 

27. y = (a + bx m ) n . Ans. -j- = z?z/z£ (a + A*" 1 )"- 1 *™- 1 . 

1 — x dy 1 -\r x 

28. y = • Ans. -f = - ? — - — — • 

Vi + a^ *& (1 + .r 2 )* 

\la + x A dy V# ( V* — Vtf) 

29. y= .__._. - ^^. -f = ^ ■ V n S n . 

Va + Vx dx 2 VxVa + x(Va+ Vx) 2 

30. 7 ='( 2 ^ + ^) V^ ^ + *J*' 

dy 4 V^ + 3 V.*: 

^*r. -Z = — z_ — ^ . 

dx 4 v» V ^ + v* 



Differentiation 29 



TRANSCENDENTAL FUNCTIONS. 
THE LOGARITHMIC FUNCTION. 

30. The differential of the logarithm of a variable is equal to 
the modulus of the system into the differential of the variable divided 
by the variable. 

Let u be the variable, and let m be the modulus of a system 
of logarithms whose base is a. 

To prove d (log a u) = m 

Let c be any constant and let 

u = cv (a) 

Differentiating, du = cdv ; 

or, du = - dv. 

v 

Hence — = — (b) 

u v 

Applying logarithms to (a) 

\og a u = \og a v + \og a e; 

Differentiating, d (log a u) — d (log a v) .- (V) 

Dividing (V) by (b). we have 

^(log a «) <t(log a v) 



du dv 

U 7) 



(d) 



Now let us consider v at some one of its values, say v r . 
Then the ratio d(\og a v) 



dv 

v 



= m, 



30 Differential Calculus 

where m is some constant. When v = v r we have from (a) 
u' = cv' , i.e., some particular value of u. Since the ratios in 
(d) are always equal, we also have, 



du 
u 



= m. 
u' = cv' 



But c is any constant, .*. u' (= cv rs ) is any value of u, .'. , gen- 
erally, 

<?(}og a u) 



da 
u 



= m. 



da 
Hence, d(Jog a u) = m — (10) 

Since the constant m depends only upon the constant a for 
its value, it is the Modulus of the system of logarithms whose 
base is a. 

Cor. In the Napierian system the modulus is unity. Repre- 
senting the base of that system by e, we have, 

d(log e u) = ~ (11) 

Hence, The differential of the Napierian logarithm of a variable 
is equal to the differential of the variable divided by the variable. 

31. Remark. — The simplicity of equation (11) as compared 
with equation (10) explains the reason for the almost exclusive 
use of the Napierian system of logarithms in the higher analysis. 
We shall therefore restrict our attention to this system in the 
investigations which follow, unless the contrary is expressly in- 
dicated by the use of some subscript to the logarithmic symbol. 

32. By means of the formula derived in article 30 for the dif- 
ferential of a logarithm we can now show that d(u n ) = nu n ~ 1 du 
when n is incommensurable. See § 27, Case 4. 



Differentiation 31 

Let y = u n , 

in which n is any incommensurable number. 
Applying logarithms, log y — 11 log u ; 

dy du 
y u 

d(u n ) du 

i.e., — ^ = n 

u n u 

Hence, d(ii n ) = nu n ~ x du. 

Equation (9) is therefore true in all cases. The above meth- 
od is of course applicable whatever the value of ;/. 

THE EXPONENTIAL FUNCTIONS. 

33. To prove 

d(ii v ) = vu v ~ 1 du + u v \ogudv, 

in which u and v are variables. 

Let y = u v ; then 

logy = v log u. 

dy du , . 

.*. — = 7) h log «/7Z/, 

y u 

d(u v ) du . 

i.e., —^~ = v (- logudv; 

u u 

hence d(u v ) = e otC~ 1 du + u v \o% udv . . . . (12) 

Cor. 1. Let v = n = a constant; then formula (12) gives 
directly and generally 

d (u n ) = nu 1l ~ 1 du, 

as previously determined. See §§ 27, 32. 

Cor. 2. Let ^ = a = a positive constant; then formula (12) 
becomes 

d(a v ) =a v log ad<v ...... (13) 



32 Differential Calculus 

Hence, the differential of a constant affected with a variable ex- 
ponent is equal to the consta?it affected with the same exponent into 
the logarithm of the constant into the differential of the exponent. 

Cor. 3. If u = e = base of the Napierian system, we have 
(since log e = 1), 

d(e v )=e v d<v. (14) 

If we compare formulas (9) and (13) with formula (12) above, 
we see that the following rule may be given for the differential 
of a variable affected with a variable exponent : 

The differential of a variable affected with a variable expo?ient is 
equal to the swn of the results obtained by differe?itiati?ig, consider- 
ing first one variable and then the other to be a constant. 

34. Formulas. 

. %£tj d(hgav) = m 

d{u v ) = <vu v - x du + u v log ud<v. 
d(a v )=a v log adv. 
d(e v ) = e v dv. 



EXAMPLES. 

Differentiate : Ans. r 

2 v 
1. logjc 2 . d(\ogx 2 ) = -dx^sr^ 

X '■■: Jt 

a ■+- x 2 



2. log (3 ax + x s ). d(\og( 3 axj- x s )J= ^K^ dx,. 

3. y = log Vi — x z . dy = - -^ .dx. . — , 5 

2 * 3 — I , ; J: " 

4. y = x™*?. dy = x m = x iin + x^dxr 



+ t 



ii 



J ■•': * 



Differentiation 



33 



5. y = (x 2 — 2 x -f- 2) e*. dy = x?er°dx. 

6. y = e* (1 — x s ). dy = (1 — 3 jc 2 — .x 3 ) ^dx. 



20. j^^log*. 



dfc 



7. y = log (log A*). 


x iogx 




8. y = x log x. 


dy = (1 -{- log a;) da:. 




9. _y = .xV. 


•/y = a x (n + •* log tf) # M ~ Va;. 




10. y = \og a x s . 


7 3 m 1 
dy = dx. 

X 




1 1; 11 — 


(1 Hf^E 5 ^ 




*^ I + X 




12. jy = ^log*. 


//>> = <f r ( — h log a: 1 dfof. 


"" 


13. y fc= #*. 


</j/ = (log at + 1 ) x^dx. 




14. ^ = at"^. 


dy = log # (log x -\- 1) + - x^x^dx. 




1 

15. jy == A*. 


1-2 

</y = (1 — log x) x x dx. 


| 


16. jy =« 6 "k 


dy = ye* log adx. 




17.-^ = ^. 


dy = ye^dx. 


^ * 


18. y = e x ° C . 


(fy = yx* (log x 4- 1) ^p. 




»■-©•-. 


^ = ^ ( lo ^ - ') <**• 





*/y == e™* I — \- m log x ) d&c. 



21. y = log^A*. log (log a:) — log ^. 



</y = 



log (log x) dx 



x 



34 Differential Calculus 

22. y = (# 3 — 3 x 2 + 6 x — 6) ^. ^ = x z e°dx. 

23. 7 = VS - log(i + \©- * = 2(l + V - } 

JV lop" X 

24. _y = log(i— ^f)H \ogx. dy = - — - — 



I — x (i — x) 2 



dx. 



25. y = \og 5 x. dy = 5 log 4 .r — 

/i -f- #\* , dx 

26. 7 = log . dy — 



1 — x 1 — x 2 



. Vi — ^ 2 4- x ^[2 _ rtk V2 
27. jy =log . dy — T ——= 

Vi-* 2 (Vi-^+^V2)(i -x 2 ) 



dx 



28. y = log( V*+w + \lx — n). dy = , * 

'2 y (x -\- m) (x — n) 

It frequently happens that the process of differentiation of 
algebraic functions may be greatly simplified by applying loga- 
rithms before beginning the operation. Thus, 

x m ' 

30. y = , x ro •'• tog-)" = ** lo g* — m lo g C 1 + x ) 
^1 -f- .#y 

dty { dx dx 

= m 



y I x 1 -\- x 

dx 



= m 



x(i -\- x) 



. dx 

dy =my — — - — r 
x(i -\- X) 

x m — * 

= m -, " x ^_i dx. 

(1 + x) m + 1 



Differentiation 



35 



Solve the following by this process : 

Ans. 



31. y = 



x 



I -J- X 

32. y = (i + m x ) z . 

(i -|- x 2 )x 

33. y = v J - 

Vi -** 

J(x + df 

34. y = V -. 

V x — a 



dy = 



dx 



(i + *) 2 ' 
*/y = 2 (i -\-m x ) m x log wdk. 
1 + 3 ^ 2 



</y = 



V(i-* 2 ) 



2 x* 



,2\3 



^/X. 



^ = (^- 2 «)V ( ^±^- 3 ^. 



THE TRIGONOMETRIC FUNCTIONS. 

35. The differential of the sine of an angle is equal to the cosine 

of the angle into the differential of ^_ ^_ t 

the angle. 

Let POC be any angle gen- 
erated by the line OP, taken as 
the linear unit, revolving upward 
about O as an axis ; then, in 
circular measure, 

Length of PC = u = measure of POC. 

If length u becomes a uniformly changing variable at the in- 
stant the generating point reaches the position P, then, § 18, 

PT = du and Z>T= d sin u 
(since AP = sin u). From the right triangle DTP, we have, 




i.e., 



Z>T=PTcosDTP, 
d (sin u) = cos a du ...... (15) 



36 Differential Calculus 

36. The differential of the cosine of an angle is equal to minus 
the sine of the angle into the differential of the a?igle. 

From the right triangle DTP, Fig. 4,* we have 

DP = FT sin DTP, 
i.e., — d cos u = sin u du, 
or d (cos u) = — sin udu (16) 

since OA = cos u and OA is a decreasing variable, § (8), II. 
Otherwise, thus : let u = u in equa. (15), then 

d sin f u ) = cos ( u\d[ u 



2 ) \2 J \2 

i.e., d (cos u) = — sin u du. 



37. The differential of the tangent of an angle is equal to the 
square of the secant of the angle into the differential of the angle. 



To 


prove 














•/(tan u) 


= 


sec 2 u 


du. 


From 


trigonometry, 












tan u 




sin u 





cos u 

Differentiating, 

, , N cos u ("cos u du) — sin u ( — sin u du) 

*/(tan u) = i <-. — , ^ '- 

' cos J u 

(cos 2 u + sin 2 u) du 
cos 2 u 

= — r~ du > 

COS J u 

.-. </(tan tt)= sec 2 udu (17) 



Differentiation 37 

38. The differential of the cotangent of an angle is equal to minus 
the square of the cosecant into the differential of the angle. 

To prove 

d (cot u) = — esc 2 u du. 

We know that cot u 



tan u 

d (tan u) 



tan 2 u 
sec 2 u 



.'. d (cot u) = — 

tan 
<;p r 2 u. 

du, 
tair u 

.'. d(cot u) =— esc 2 u du (18) 

Otherwise, thus ; let u = ( u\. Substituting in equation 

(17), we have 

d tan ( u\ = sec 2 ( u ) d ( - 

i.e., d (cot u) = — esc 2 u du. 

The complementary functions which follow may be differen- 
tiated by the student in the same way. 

39. The differential of the secant ^of an angle is equal to the 
product of the secant, the tangent and the differential of the angle. 

To prove 

d (sec u) = sec u tan u du. 



We know that 


1 


cos u 


- 


d(cosu) 
.-. </ (sec u) = — - — - 

COS J u 




(— sin udu) 




cos 2 u 




.*. cf(sec u) = sec u tan udu 



(19) 



38 



Differential Calculus 



40. The differential of the cosecant of an angle is equal to minus the 
product of the cosecant, the cotangent and the differential of the angle. 

To prove 

d (esc u) = — esc u cot u du, 
i 



We know that 



esc u = 



d(csai) = — 



sm u 

cos udu 



sin' u 
= — cscu cot udu 



(20) 



41. The differential of the versine of an angle is equal to the 
sine of the angle into the differential of the angle. 

To prove 

d (vers u) = sin u du. 

We know that vers u = i — cos u, 

,\ d (vers u) = — d cos u 

= sin u da (21) 

42. The differential of the coversine of an angle is equal to minus 
the cosine of the angle into the differential of the angle. 

To prove 

d (covers u) = — cos u du. 

covers u = 1 — sin u, 
d (covers u) = — cos udu .... (22) 



We know that 



43. Formulas. 



c/sin # = cos udu 
d cos a = — sin tfcfo 
cftan a = sec 2 a</tf 
dcot u = — csc 2 u du 
dscc u = sec a tan acfa 
(/esc t* = — esc u cot #</(/ 
(/vers t* = sin c*(/ff 
(/ covers a = — cos u du 



Differentiation 39 

EXAMPLES. 

1. y = sin 3 x. dy — 3 cos 3 xdx. 

2. y = sin 3 3 x. dy = 9 sin 2 3 ^ cos 3 jc^/r. 

3. jj; = cos mx. dy = — m sin mxdx. 

4. y = tan 2 5 #. */y = 10 tan 5 x sec 2 5 xdx. 

5. jy = cot x 2 . dy = — 2xcsc 2 x 2 dx. 

6. y = sec 4 x dy = 4 sec 4 # tan 4 ^:^r. 

7. jy = sec 2 #Jt\ dy = 2 n sec 2 nx tan nxdx. 

8. jy = log sin #. ^v = cot xdx. 



9. y = x 



_ /sin 3: \ 

#y = ^ sma f (- cos # log x 1 d*. 



tan x — tan d .# 

*- 10. y = -. . dy = cos a. xdx. 

J sec 4 * J 

• /1 v 7 cos (log x) _ 

11. j = sin (log *). tfy = — - — - ax. 

dy 

12. y = esc 11 x. — = — ncsc n x cot #. 
^ //# 

13. j = sin (1 -f- x 2 ). — - = 2 * cos (1 ■+- x 2 ). 

14. jy = sin (sin at). — = cos # cos (sin x). 

15. _>> = cos mx cos «*. 

dy . . . 

-— z= — (m cos /at sin ;;/* -(- ;/ cos /;z>r sin ;/*). 

16. If sin 2 # = 2 sin # cos x prove by differentiation that cos 

2 x = cos 2 x — sin 2 #, and conversely. 

2 tan a? 

17. If 2 sin at cos x = — - prove bv differentiation that 

1 + tan 2 x * y 

cos 2 x — sin 2 x = — , and conversely. 

1 + tan 2 x' J 



■-% 



40 Differential Calculus 

18. If sin (x -\- y) = sin x cos y -f- cos x sin jy "prove by differ- 

entiation that cos (x -\- y) = cos x cosy — sin x sin y, and 
conversely. 

19. If sin (x — y) = sin x cos y — cos x s'my prove by differ- 

entiation that cos (x — y) = cos a: cos y + sin .# sin y, and 
conversely. 



T . . . . /i — COS X , 

20. If sin ^-.r = y prove that cos fx = 1/ 



i -f- cos x 



THE CIRCULAR FUNCTIONS. 

44. 77/<? differential of an angle is equal to the differential of its 
sine divided by the square root of one minus the square of the sine. 

To prove 

js - -l \ du 

//(sin y u) = 



\FT^~u< 

Let v = sin~ 1 u. Then sin v = u, 

.'. cos vdv = du, 

.♦. </?/ = d(sm~ l u) 



COS 27 ^ 

but cos v = Vi — sin 2 v = yi.^0^ 

Hence d(sm~ l u) = . .... (23) 

The student may deduce this as well as all the following 
formulae by solving the formulae for the differential of the trigo- 
nometric functions for du. Thus *~ 

Art. (35), Equa. (15), d (sin u) = cos u du, 

d (sin u) d '(sin ar) 

,\ du = — - = . . ^-= . 

cos « Vi - sin 2 u 



^ 



v 

Differentiation 41 

45. The differential of an angle is equal to minus the differential 
of its cosine divided by ihe square root of o?ie minus the square of 
the cosine. 

7/ i x du 

To prove d (cos 1 u) = . 

Vi — u 2 
Let v = cos -1 u. Then cos v = u, 

.'. — sin vdv = du, 
.'. dv = d (cos -1 it) = 



sin v 

but sin v = Vi' — cos 2 v = Vi — «/ 2 . 
Hence ^(cos" 1 **)^ ■ • • ( 2 4) 

46. 77^ differe?itial of an angle is equal to the differential of 
its tangent divided by o?ie plus the square of the tangent. 

7/ _i s du 

To prove d (tan x u) = — ■ — - • 

r ' i -f- u* 

Let v = tan -1 u. Then tan v = u, 

.'. sec 2 vdv = du, 

7/ _in du 

.-. dv = d (tan * u) = — — ; 

sec 5 v 

but sec 2 z/ = i + tan 2 v = i -f- w 2 . 

Hence </ (tan~ * a ) = ^— , — = ( 2 5) 

I -hit 

47. The differential of an angle is equal to minus the differential 
of its cotangent divided by one plus the squa?-e of the cotangent. 

_ 7/ _i n du 

To prove d (cot * u) = : — o • 

r v J i -h zr 

Let v = cot -1 z/. Then cot v = it, 

.'. — esc 2 vdv = du, 

du 



.-. dv — d (cot 1 it) = ■ 

v y CSC* V 



.2 



42 Differential Calculus 

But esc 2 v = i + cot 2 v = i 4- ?' 2 . 

Hence d(cot~ 1 a) = - T -^ r .... (26) 

48. 77/<? differe7itial of an angle is equal to the differefitial of its 
secant divided by the secant into the square root of the square of the 
secant minus one. 

Let v = sec -1 u. Then sec v = u. 

Hence dis^ u) = - f U (27) 

The exercise is left for the student. 

49. The differential of an angle is equal to minus the differential 
of its cosecant divided by the cosecant into the square root of the 
square of the cosecant minus o?ie. 

To prove d (csc~ x u) — . ( 28 ) 

The exercise is left for the student. 

50. The differential of an angle is equal to the differential of its 
7>ersi7ie divided by the square root of twice the versine minus the 
square of the versine. 

To prove d(vets~ 1 u)= . = = ( 2 9) 

v 2a —tr 

The exercise is left for the student. 



51. The differential of an angle is equal to minus the differential 
of its coversine divided by the square root of tivice the coversine 
minus the square of the coversine. 

To prove (/(covers -1 u) = , (3°) 

V2ff-t* 2 

The exercise is left for the student. 



Differentiation 



43 



52. Formulas. 



d (sin" 

d (cos" 

</(tan" 

</(cot" 

d (sec" 

</ (esc" 

(vers" 

d (covers" 



«) = 



du 






Vj-a 2 



<fc 



J -ha 2 



J -ha 2 
c/a 



tr) = - 

*) 

u) 



du 



da 



V2 a - V 



V2a-a 2 



EXAMPLES 



1. jy = sin 1 (3# — i). 



2. y = sin" 






dy = 



3 /£r 



V6 x — g x 2 
dx 



Z, y — x sin 1 ^. 

4. j = tan -1 

5. ^ = tan -1 



Vtf 2 - a: 2 

*/y = [sin -1 aH — )dx. 

\ Vi-W 



# 



\ll - X 2 

2 ^C 



-X 2 






^r 



Vi -^ 

2 rtfo 

I +^ 2 



44 Differential Calculus 



6. y 



. 2 x , 2(1— x 2 ) dx 

^tan" 1 -• dy = v , 9 J — .- 

1 + jr 1 + 6 .ar -+- x* 

_ 1 a */x 



7. y =cos x - • dy - 



a 2 —x 2 

x 2 ' 1 — 1 2 ^r n_1 , 

8. j = cos x -r-— - tfy = dx • 

^ X 2 ' 1 + I ^ ^ n + I 

^1 , dx 

9. y = sec x wx. #y = - . 

•* V;;/ 2 ^ 2 — 1 

1 ^ 7 dx 

10. 7 = sec x — — =- • dy = — == • 

\la 2 — x 2 ^Ja 2 — x 2 

, [—• 7 Vi + cscr . 

11. y = sin l V sm .r. </y == dx • 

. 8 ^ 2 ^r 

12. y = vers x — • dy = — — • 

9 V9 a" — 4X 2 

/ o 1 x dy ,x 

13. y — (a 2 + jv 2 ) tan -1 - • -f- = 2 * tan -1 - + 

y a dx a 

14. v = (* + a) tan - x — — — \fax. -4- — tan -1 V/ - • 

v 7 v« dx * a 



a. 



1 ^ — e x dy 

15. y = cos -1 — 



2 

x 



16. jy 



,,tan ^ x 



e° -\- e x dx e* -\- e 

dy __ e tan ~ lx 
dx 1 -j- x 2 



. x dy 1 

17. y = ,sm~ 1 - . -j- 



18. j=sm- 1 — — • -f 



Vi + jp 2 </# I + ** 

+ 1 ^/v 1 

V2 dx Vi -2X-^ 

.1 dy 2 

19. jy = sec" 1 — -^ f- = • 

2 XT — I ^%" Vl — X 2 

20. y = x^la 2 - x 2 + a 2 sin" 1 - . f = 2V« 2 -^. 

a dx 



Differentiation 45 



00 i i _ _ 4 j x — a dy 2 ax 2 

V x 4- a 



21. y — tan -1 - + log . 

a y x -\- 

22. y = log tan ^ + ^j 

23. _y = log tan -1 x. 



24. ^ = tan- 1 ^ r -f = — £- 5 

i — 3 jr # * i -f- ** 

i i dy i 

25. y = tan -1 * H 



dx 


4 4 






dy_ = 

dx 


sec *. 






dy 


i 






dx 


(i + **) 


tan 


"- 1 * 


dy 


3 







d* .x 2 (i + x 2 ) 

i 5 , dy x sin -1 x 

26. y = x — Vi — x 2 sin -1 *. -7- = — t 

//* Vi — ** 

27. _y = (2 * 2 — 1) sin -1 x + * Vi — x 2 . 



dy • _i 

-j- = 4 * sm * *. 
#* 



28. y = tan x sin 1 x. 

dy 



dx (1 + (sin J *) 2 ) Vi — x 2 

29. y = x 2 + (sin -1 *) 2 — 2 sin -1 x . x Vi — x 2 . 

dy 4.x 2 sin -1 * 



30. y = tan 



-1 



doc Vi - * 2 

Vi — cos* dy 1 

Vi + cos* dx 2 



, 1 — cos * dy 

31. v = log \/ • — = esc*. 

T I + COS * tf* 

1 3 + S cos x dy 1 

32. jy = cos -1 ° 



33. y 



5 -f- 3 cos * dx 5 + 3 cos * 

(1 — x 2 )? sin -1 * 



* 

dy 1 — x 2 1+2* 2 



(1 — * 2 )* sin 1 *. 



46 Differential Calculus 



VI — X^2 + X 2 . X VI 

34. y = log * == + tan-* ^_^ . 

yi +.rV2 + ** 1 — ar 

ax j _j_ ^4 

1 -f- •* 1, \ + x -\- x*" ,— ,Jtr V? 

35. y = \og — i_ + -log— ^_+V 3 tan- 1 *- 

I — Jf 2 I — X + X* I — XT 

dy 6 

dfc 1 — x 6 



Limits 47 



CHAPTER IV. 



LIMITS. 

History. — What is known as the " method of limits " in the Calculus 
is founded on the following lemma in the first book of Newton's Principia 
(1687): 

" Quantities and ratios of quantities, which in any finite time converge 
continually to equality, and before the end of that time approach nearer the 
one to the other than by any given difference, become ultimately equal." 

53. In deriving the differential formulae in Chapter III., we 
have taken as the basis of the operation what is known among 
mathematicians as the " Method of Rates." We shall consider 
in this chapter another method, known as the " Method of 
Limits," and show how by this method all the foregoing differ- 
ential forms may be derived. 

54. Limit. The limit of a va?'iable is a fixed value fro?n which 
it can be made to differ by less than any assignable quantity but 
which it never reaches. Thus, 

Limit of x = Limit (.66666 . . .) = §, as the figure 6 is an- 
nexed an indefinite number of times. 

Limit of x = Limit (1 + i + i + i + tV + • • •) = 2 ' as 
the number of terms indefinitely increase. 

The limits of the area and of the perimeter of a regular in- 
scribed polygon are respectively, the area and circumference of 
the circumscribing circle as the number of sides of the polygon 
indefinitely increase. 

sin x 

Limit = 1, as the value of x indefinitely diminishes, 

tan x 

i.e., as x approaches the value zero. 



------ 



48 Differential Calculus 



Cor. It is evident from the definition that the difference be- 
twee?i a variable a?id its limit is a variable whose limit is zero. 

55. The student should be careful to distinguish the limit of 
a variable as above defined, from the term limit ^as "Ordinalrily 




used. Thus in the circle x 2 + y 2 =.<£, or y = ± Y&y— x 2 , we 
are accustomed to say that the limws\oi the, values of x are 
± a. By the term limit as thus iised we mea^.hat Beyond 
these values there are no corresponding values of y, i.e., there 
are no points on the locus. 

56. Principles. I. If two variables are always equal and each 
app?'oaches a limit, their limits are equal. 

Let u = v, and let limit of u — a and limit of v = b ; since 
// = v, we have a — u = a — v. But a is the limit of u ; hence 
a is also the limit of v, §54, Cor. But b is the limit of v\ 
hence a = b. 

' - ■ . s 

II. The limit of the sum of any number of variables is the sum 

of their limits. 

Let ?/, v, w y . .--.. be any number of variables whose limits 
are a, b, c, . . *% respectively, then 

(a — u) + (b — v) + (c — w) + . . . , 
or (a + b + c +...) — (u + v + w +.. .) 

is a quantity whose, limit is zero. § 54, Cor. 

Hence Limit (u ■+ v + w \ . .) = (a -{- b -\- c . . .). 

III. The limit of theSproduct of any number of variables is the 
product of their limits. 

Let u tnd v be variables whose limits are a, and b, respectively. 

Let > a — u = x, and b — v = y, . 

in which x and y are variables whose limits are zero: § 54. 



A 



*tffc 



Limits 49 

Hence u = a + x, v = b -f- y, 

hence nv = ab + bx 4- ay + xy ; 

«z> — dr£ = &c + ay + ay. 
But the limit of the second member of this equation is zero ; 

Hence Limit (tiv) = ab. 

And so for any number of variables. 

57. Notation. In order to express the fact that a given func- 
tion approaches a certain limit as the variable which enters it 
approaches a certain other limit, it is convenient to adopt some 
form of notation. The form in common use is illustrated by 
the following : 

_ . . Tsin x] 

Limit = i. 

L X Ja- =0 

This expression is equivalent to the sentence ' the limit of 

sin x . .,..., 

as x approaches zero as its limit, is i. 



i x 



58. The Differential Coefficient or First Derivative of a function 
is the limit of the ?-atio of the increment of the function to the incre- 
ment of its variable, as the increment of the variable approaches zero 
as its limit. 

Thus let y =f(x) be any function of x, and let A.* be an in- 
crement of x and Ay be the corresponding increment of the 

dy 
function. Let the complex symbol — represent the differential 

coefficient, or first derivative of y =f(x) ; then, by definition 

Limit 



Ay 
Ax 



dy 

Sx=o ax 



/ 



50 



Differential Calculus 



59. Remark. — The student should carefully observe that 

dy 
the symbol — as here used is a symbol representing the limiting 

value of a ratio] It is not therefore a fraction, the numerator 

and denominator being dy and dx, respectively. We shall show 

dy 
in a subsequent article, see § 68, that the fraction — is equal 

~ Ay* 

Ax 



to Limit 



Aa:=0 



s 



EXAMPLES. 
Find the first derivative of the following : 
1. y = x 2 . 

Let y = value of y when x = x -f- Ax ; that is, when x has 
taken on the increment Ax ; then 

y = (x -pAx) 2 = x 2 + 2xAx -|- Ax 2 , 
.'. j/ — y = (2 x + Ax) Ax ; 

But j/ — 7 = Aj>, 

Ajj> 



Hence, Limit 



Ax 



i.e. 



Ax 

Ax=0 

dx 



= 2 x + Ax. 






= 2 x. 



2. 7 = x 3 + 3. 

.•; y' — (x + Ax) 3 + 3 = x 3 + 3 x*Ax + 3XAX 2 + Ax 3 + 3, 

•'• y' ~ y — ^y — 3 -^^x + 3 ^Ax 2 + Ax 3 , ^ 

Aj 



Ax 



= 3 x 2 + 3 xAx + Ax 2 



Hence, 



i.e., 



Limit 



'Ay 

Ax 



Ax= 

h 

dx 



= 3X 2 . 



Limits 



51 



£» 



3. y ~== (2 x — i) (x -\- 2) 
= %!& + 3^-2, 
.•. £gf = 2 (x + Ax) 2 -j- 3 (jc + Ajc) — 2 



s ;rt 



-2 x 2 -\- 3 x — 2 -\- 4.x Ax -f- 3 A-* + 2 A.* 2 , 



•■• Ar = j£ — y ^M/4 & £3 + 2 A *) A *> 



Ay 



Ax 



= 4^ + 3 + 2 Ax. 



Hence, 

i.e., 
4. 



Limit 



'Ay 
Ax 



= 4^ + 3; 

Ax=0 

dy 



m 

y = ^ 



J' 



. 7 = 

Ay = 

Ay 

Ax 



m 



(x -\- Ax) 2 



m 



(x -\- Ax) 2 x 2 
2 x-\- Ax 



tn (2 x + A -*) Ax 

J? (x + A.*) 2 



— w 



Hence, 



.r 2 (x + A.*) 2 

T . • r A /i 

Limit — - = — m — - 

L a ^Jax= 



X* 



dy 
dx 



2 m 
x 3 



5. y = 



# + 



2 ^ 
6. y = \lx? +a. 

1. y = a V^. 

8. >.= 



^/jc 2 x 2 

dy x 

dx 2 



m 



V^ + 



dy 
dx 



mx 



^(x 2 + 2) 3 



52 



Differential Calculus 



9. y = 



a -\- bx -\- ex 1 



x 



dx 






x 



10. y =/(*). 



^ T • • 

— - = Limit 
ax 



us? 



+ Ax) -/(*)" 



Ax 



|Az=0 



60. The foregoing examples illustrate the meaning of the 
term ' differential coefficient ' and explain the process by which 
it may be derived in any given case. The process, however, is 
lengthy and tedious, and, in a large majority of cases, very diffi- 
cult. In the practical application of the i method of limits ' to 
the derivation of differential coefficients it is usual to derive a 
system of rules by aid of which the operation is greatly simpli- 
fied. These rules have been derived already by the principles 
of the ' method of rates ' (Chapter III). We shall now show 
how they may be derived by the ' method of limits.' 



61. To prove 



d (u -f- v) du dv 
dx dx dx 



Let y = u -\- v, in which y, u and v are functions of x, and 
let y = y', u — ?/, v = v' when x = x -f- A.r. 

Then y f = it + v', 

and Ay = y' — y = ?/ + v' — (u + v) 

= u' — u -f- (v f — v) ; 
Ay = Au + Av. 



i.e. 



Hence 

Therefore § 56, II. 

"Ay" 



Ay Au Av 
Ax Ax Ax 



Limit 



i.e., § 58, 



Ax 



= Limit 



Ax=0 



'Au 

Ax 



+ Limit 



Ax=0 

dv 



L a ^Ja^=o' 



dy du 

dx dx dx 



Limits 53 

Replacing y in the first member by its value u -f- v, we have 

diu +v) du dv 

= jZ+sI.' Compare §(24), 2. 









dx dx ' dx 


62. 


To 


prove 


d(uv) dv 
dx dx 


Let 






y = uv\ 


then, 






y = u'v\ 


hence, 






y' —y = u'v' — uv 



du 
dx 



Adding and subtracting uv in the second member, we have 

y' —y = u (v r — v) + v (u r — u) ; 

that is, Ay = u Av + s/ A?/ . 

A_y Av , A ^ 

A.r Ax Ax 

Hence, § 56, 11., in., 

Limit lx^l = Limit «-A\ . + Limit KA^ ; 

djy //z> du 

dx dx dx 

since v is the limit of v as A:r approaches the limit zero. 
Hence, since y = uv, 

d (uv) dv du _ » / \ 

A "^fc+'ZT Compare § (25), 3. 



63. To prove 



\ zy ^r ^r 

dk v 2 



54 Differential Calculus 



Let 
en, 


u 

y = -\ 

V 

, u' 

y = 7 ; 

, u' u 

■■■» -y = ^-- v 

u'v — v'u 






v'v 






v {11' — u) — u (v' - 


-V) 




v'v 




Hence, 


v Au — u Av 

^y - . 

V V 

Ay v Au u Av m 
Ax v'v Ax v'v Ax 





.'. § 56, II., III., 

_ . . \~Ayl T . . Y v Au~\ T . . f u Av\ 

Limit -r- = Limit X—r- -— — Limit \-j- -— • 

L a ^Jax=o \yvAx\^ x=(i L 7;z/A ^Jax=o 

dy v du u dv t 

*' dx v 2 dx v 2 dx 

(since limit of 1/ as Ax approaches zero is v) 

,(u\ du dv 

\ v 1 dx dx 
or j = Compare § (26), 6. 





dx 


V* 


64. To prove 








d(u n ) 
dx 


. du 

nu n ~ L — 

dx 



in which n is a positive integer. 

Let j> = & w ; 

then, y f = u' n ; 

.*. Ay = u' n — u n 

= (u'-u)(u' n - 1 + u' n - 2 u+u' n ~ 3 u 2 + . . . a"" 1 ) 



Limits 55 

... ^? = — (u'"- 1 + u n -*u + i/— V -f . . . ^ w - 1 ). 

Ax Ai^ 

Hence, § 56, in., 

Limit — = Limit u rn ~ x — + Limit u' n ~ 2 u-^ 

LA*jA;c = o L a *Jax=o L a ^Ja^ = o 

+ . . . to n terms. 

dy .du . .du 

.-. -f- = u n ~ Y — -\- u n ~ x — + ... to » terms, 
dx dx dx 

since the limit of u r as Ax approaches zero as a limit is u. 

d(u n ) . du 





1^^, 






dx 


65. 


To 


prove 










dx 


.«) 


m du 
u dx 


Let 






7 


= log a u ; 


then, 






/ 


= log„«' ; 



</# 



Compare § (27), 9. 



u 

^y = i°g<y - iog a ?/ = iog a - 



/« -f- A#\ 

= log "HH 



(Au Au 2 Au 3 
u 2 u 2 2> u * 



" / 



m . / A« A?/ 2 
= — A?4i 
u 



Ay mAu 
Ax u Ax 



( Au Au 2 \ 

1 +— - • • • ); 

( Au Au 2 \ 

1 — — H 2 — • • • )• 

\ 2U $U A J 



56 Differential Calculus 

Hence, § 56, in., 

]_ . . V m Au~\ 
= Limit — — - X 

Ax=0 \_u AxJ Aa;= io 



Limit [g 



_ . . |7 A# Au 2 

Limit 1 1 r — .... 

L\ u u 2. 



Ax=0 



i.e., 



dy d(\og a u) m die _ _ , N 

Tx = *H " « S ' Compare § (3o) ' IO> 



Since the limit of Au as Ax approaches zero as its limit is 



zero. 



66. To prove 



d (sin u) dn 

— —. = cos u — 

ax dx 



Let y = sin u ; 

then, y = sin u' ; 

.-. A_y = sin u r — sin z/ 

= 2 cos ^ (2/ + u) sin ^ (#' — u) 



I Au\ . A11 ._,. , , . x 

= 2 cos ( // -+- — 1 sin — ; (Since u = u -f- A?/) 



.. Az* 
. , A v sm — A 

Aj / AzA 2 Az/ 

-— = cos u -\ 

Ax V 2 I Au Ax 



Hence, § 56, hi., 



Limit -^- = Limit cos ( u + — J 

|_A^J Aa;= io L V 2 / J^ 



Limit 



Au- 



sin 



An 
2 J 



X Limit — 



A«l 



A*= 



i.e., 



</y d(sm u) du _, c , N 

-f- == — i— — ^cosz/ — . Compare §(35), 15, 

/wr dx dx 



Limits 



57 



Since Limit 



r • A^n 
sin — 
2 



Au 

L 2 J Aj;=0 



= I. 



67. To prove 



d(sm 1 u) 

dx ^y I 






ir 



Compare § (44), 23. 



Let 

then, 



sin x u = v, or u — sin ^ ; 



du 

dv dx 



du 

dx 



i.e., 



' dx cosz/ ^j 

du 
d(s\xi~ 1 u) dx 



sin^ z/ 



^r 



Vi - u 2 ' 



68. The limiting value of the ratio of the increment of a function 
to the correspondifig increment of the variable, as the increment of 
the variable approaches 
zero as its limit, is the 
ratio of the differential of 
the function to the differ- 
ential of the variable. 

Letjy =f(x) be the 
equation of the locus 
AB, Fig. 5. Let PP' 
= As and the length 
AP = s ; then 



Y 






B . 






py 






/ i T ^~ 


A- 


I 


~>/ / <^' 


B. IB 


8 


C J 


1 


M' 



Fig. 5- 



PP = Ax = corresponding increment of x (OM), and 
RP' = Ay = corresponding increment of y {MP). 



58 Differential Calculus 

Draw the secant PP' and the tangent TS at P\ 
then 



^ = tan P'PR. 

Ax 



If we now suppose Ax to approach zero as its limit the point 
P' will approach P and the secant PP' will approach the 
tangent TS\ hence 



Limit 



"4/ 

Ax 



= Limit [tan P'PP] = tan TSX. 

Ax=0 A;r=0 



But § 18, Cor., the ratio of dy to dx is equal to tan TSX, i.e., 






tan TSX. 



Hence, Limit -— = — . 

\_Ax_\ Ax=0 dx 



Analytical Applications 59 



CHAPTER V. 

ANALYTICAL APPLICATIONS. 

1. (a) Compare the rates of the ordinate and abscissa of the 
generating point of a circle whose radius is 5. (J?) What does 
the ratio of the rates become at the points (—3, 4), (o, 5), 
(5, o) ? (c) If the ordinate is increasing at the rate of 8 feet 
a second at the point whose abscissa is 4, what is the rate 
of the abscissa and what the velocity of the point in its path ? 
(d) In which angles is the ordinate an increasing function of x ? 

(a) Referring the circle to its center and axes, we have 
x 2 +y = 25 for its equation ; 

/. 2 xdx •+• 2 ydy = o, 

x 
.*. dy = dx, 

y 

x 
i.e., the ordinate changes times as fast as the abscissa. 

(*) At the point (-3, 4), | = ~= J; at (5, o), % = ~\ 

, n dy ° 

= -oo ; at(o, 5 ),-=--=o. 

(^r) The ordinates corresponding to x = 4 are found from the 
equation of the circle to be y = ± 3, .*. since dy = 8 ft. a 
second by hypothesis, we have 

8 = — ■ — dx = =F - dx, 
±3 3 

,\ dx = =f= 6 feet a second. 



60 Differential Calculus 

From § (18), 3, we have ds = vdx 2 -\- dy 2 . 
Substituting values found above we have 

ds = V36 + 64 = 10 feet a second. 

x 
(d) Since dy = dx'\X. will be negative unless x and y have 

different signs, hence dy will be positive when the moving point is 
generating the second and fourth quadrants ; therefore, § (8), II., 
y is an increasing function of x in those quadrants. It is obvi- 
ously a decreasing function of x in the first and third quadrants. 

2. (a) Compare the rates of the ordinate and abscissa of the 
parabola y 2 = 2px. (b) At what point is the rate of y equal to 
the rate of jr? (c) Between what limits of x is the rate of y 
greater than that of x ? (d) Between what limits is it less ? 
(e) If the parameter of the curve is 8, and the generating point 
so moves that its abscissa is uniformly increasing at the rate of 
10 feet a second, what is the velocity of the point in its path 
and the rate of the ordinate when x = 8 ? (/) What is the 
rate of increase of the area of the parabolic segment at the 
instant that x = 8 ? 

P P • 

id) dy = -dx, i.e., the rate of y = -times rate of x. 

y y 

p ... 

(J?) By condition dy = dx, .*. 1 = — .'. y = p. Substituting m 

the equation of the parabola we find x =■ - /.( — ,/) or the 

extremity of the latus rectum is the required point. See 
" Analytic Geometry," p. 86. 

P 
(c) In the differential equation, dy = — dx, dy > dx when 

P 
p > y. But from the equation of the curve / >y when x < — , 

P 

.-. dy > dx between limits x = o and x = — . 



Analytical Applications 61 

• • P 

(d) Similarly we can prove dy < dx between limits x = — 

and x = oo . 

4 

(e) The equation becomes y 2 = 8 x ; .-. dy = - dx. By hypoth- 
esis dx = 10 feet a second, and the point whose abscissa jc = 8 
has y = S for its ordinate, .-. dy = - io = $ feet a second. Also, 

o 

ds = V^/jt 2 + dy 2 = V125 = n. -f- feet a second. 

(/) The area of a parabolic segment is z = § jey, .-. /& = § 
(xdy + j^^). Substituting values found under (<?) above, we 
have 

dz = § (40 + 80) = 80 sq. ft. a second. 

3. Compare the rates of the ordinates and abscissas in the 

X? y X^ V 2 

following curves — -f- — = 1 , — — — = 1 , xy = m. 

" 4. An elliptical metal plate is expanded by heat or pressure. 
What is the rate of change of its area when the semi-axes are 
4 and 6, and each is increasing at the rate .1 in. a second ? Let 
x and y be the semi-conjugate and semi-transverse axes, and 
let z be its area ; then z = irxy. See " Analytic Geometry," 
P- 136. 

.*. dz = 7r (xdy -|- ydx) = it (10) .1 = it sq. in. a second. 

5. Steam is admitted by a valve into a circular cylinder, one 
end of which is closed by a piston. If the diameter of the 
base of the cylinder is 1 foot, and the steam is admitted 
at the rate of 10 cu. ft. a second, at what rate is the piston 
moving ? 



62 



Differential Calculus 



Let y = volume and x = altitude of cylinder at any instant ; 
then 

.*, dx = — feet a second. 

7T 

6. Gas is introduced into a thin elastic spherical film at the 
rate of 10 cu. ft. a second. At what rate is the radius increas- 
ing when the volume is cu. ft. ? 

3 

Let y = volume and x variable radius ; then 



y = - ttx 5 , .'. dy = 4 7rx 2 dx, 
3 



.*. dx = 



io 



dy 

4 7TX 2 4OO 7T 40 7T 



feet a second. 



7. A man 6 feet in height, walking at the rate of 2 miles 
an hour, passes under an electric light 18 feet above the pave- 
ment. Assuming the pavement to be horizontal, find (a) the 
rate at which the man's shadow is lengthening ; (/?) the velocity 

B of the end of his shadow ; (c) 
the rate at which he is receding 
from the light when 15 feet 
from its foot. 

Let D be the light, and BM 
^ the position of the man at any 
Fig. 6. instant. 

(a) Let AB = y, BC '= x\ then, from similar triangles, 

y y + x 1 




hence, 



dy = -dx = 1 mile per hour. 
2 



Analytical Applications 63 

(J?) Let AC = y and BC = x ; then 

y _ y-x . v _3„. 

i8~ 6 "•-> ; - 2 *' 

hence, ^/y = - dx = 3 miles per hour. 

2 

(V) Let A be the man's position, and let AD = y and AC = x; 
then, 

.%■ I K 

hence, dy = - dx = — 2 = i4 miles per hour, nearly, 

j 24 

8. Two ships start from Sandy Hook at the same time, one 
going N. 30 E. at the rate of 10 miles an hour, the other going 
due east at the rate of 12 miles an hour. At what rate are 
they separating at the end of two hours ? 

Let y = distance of ships apart at the instant, and let // and 
v be the varying distances of the vessels from Sandy Hook ; 
then 

y 2 = u 2 -f- z' 2 — 2 iiv cos 6o° = u 2 4- i' 2 — uv ; 

.*. 2 ydy = 2 udu -f- 2 vdv — ndv — vdu ; 

(2 u — 7') du + (2 v — ?/) /^7' 

.-. dy= . = = 2 y,j miles an hour. 

2 \'ir-{- zr— uv 



>f an equilateral triangle is 2 feet, and is 

ite of 2 inches a minute. At what rate is 

the area increasing ? At what rate is the perimeter increasing ? 

Am. 32 V3 sq. in. a minute, 4 V3 in. a minute. 

10. The apothegm of a regular hexagon is 2 feet, and is 
increasing at the rate of one inch a second. At what rate is the 
area increasing? Ans. 96 V3 sq. in. a second. 



64 



Differential Calculus 



11. The altitude of a cone is constantly equal to twice the 
diameter of the base. If the altitude is 3 feet, and increasing 
one inch a second, at what rate is the volume changing ? At 
what rate is its convex surface changing ? 

T7 sq. in. a second. 



Ans. 81 7r cu. in. a second, f 



TV 



12. A reservoir in the form of an inverted conical frustum, 
radius of smaller base =100 ft. and elements inclined 45 to 
horizon, is used to supply an adjacent town with water. If the 
depth of the water at any instant is 10 feet, and is decreasing 

at the rate of two feet 
a day, at what rate is 
the town being sup- 
plied ? 

Let AC = x, OC = 
a ; then C? C' = a 4- x. 
Let y = volume ; then 




y 



7TX 



](a 4 jt) 2 + a 2 + (a + x)a)\ 



= - (3 a 2 x 4 3 ax*-\- 0?). 
3 

dy = - (3 a 1 + 6 ax + 3 x 2 ) dx 
3 
= 24200 7rcu. ft. a day. 

13. Under the action of internal forces a circular cylinder is 
changing. When the diameter is 24 in., and increasing at the 
rate of 1 in. a second, the altitude is 48 in., and decreases at the 
rate of 2 in. a second. At what rate is the volume changing? 
At what rate is the convex surface changing ? 

Ans. 288 7T cu. in. a second. Not changing. 

14. (a) Compare the rates of the ordinate and abscissa of 
the generating point of the logarithmic curve, y = \og a x. (J?) At 



\ 



Analytical Applications 65 

what point are the rates equal ? (V) How do the rates compare 
when the moving point crosses the .r-axis ? 

/ v , m . . . . . m . . . 

(a) ay = — dx, i.e., the rate of y is — times the rate ot x. 

(J?) At x = m. 

(c) When y = o, x = i, .*. ay = mdx. 



15. Which increases more rapidly, a number or its logarithm ? 

m 
Let x = number, then d log a x = — dx ; hence, if ;;z > jc the 

logarithm increases more rapidly ; if m < x the number in- 
creases more rapidly, and if m = x, the rates are the same. In 
the Napierian system m = i, .-. dx = xdlogx, i.e., the number 
increases more rapidly or more slowly than its logarithm accord- 
ing as x > or < than i. 



16. (a) How much more rapidly is the number 342 increas- 
ing than its common logarithm ? (p) How much more rapidly 
than its Napierian logarithm ? (V) If the number increases by 
1, how much will its common logarithm increase? 

(a) d log 10 x = — dx = — — — dx, 

x 342 

since m = .434295 in common system. 

342 

Hence dx = d\op^x =788 d\op*. n x, 

.434295 8 

i.e., the number increases 788 times as fast as its logarithm. 

(b) Since m = 1 in Napierian system, 

dx = 342 dlogx; 

i.e., the number increases 342 times as fas,, as its logarithm in 
the Napierian system. 



66 Differential Calculus 

(V) Since dx = i, we have 

.A-ZA2QC 

d\og 10 x = yo X i = .00126, 

34 2 

i.e., the logarithm of 343 is greater than the logarithm of 342 
by .00126. This decimal is the tabular difference correspond- 
ing to the number 342 as given in tables of common logarithms. 

17. What should be the tabular difference in tables of com- 
mon logarithms for numbers between 6342 and 6343 ? 

Am. .000068. 

18. Assuming that the ratio of the rates of change of a num- 
ber and its logarithm remains constant while the number 
changes from 245 to 245.15, find the logarithm of the latter, 
assuming the logarithm of the former to be 2.389166. 

Ans. 2.389193. 

19. Prove by logarithms that d {uvw) = uvdw + uwdv + 
vwdu. Applying logarithms we have 

log (uvzu) = log u -f- log v + log w ; 

d '{uvw) dn dv dw 

/. -^ J - = — + - + — ; 

uvw uvw 

.'. diiivw) = uvdw + uwdv + vwdu. 

'u\ vdu — udv 



20. Prove by logarithms that d\— j 



21. Show that the ratio of the rates of a x and x is equal to 
a log a when x = 1 ; of e* and x is equal to e when x = 1 ; of x°° 
and x is equal to 1 when x = 1. 

22. (a) Which increases more rapidly, an arc or its sine ? 
(b) Where are their rates of increase the same ? (V) Where is 
the rate of the arc twice that of the sine ? (d) When the arc 
is 30 what is the ratio of the rates ? 



Analytical Applications 67 

(a) d (sin x) = cos x dx. As the cosine is in general less than 
i, the rate of the arc is in general greater than the rate of the 
sine. 

Qi) By hypothesis d sin x = dx ,\ cos x = i .'. x — o, i.e., 
when the arc is zero its rate and that of its sine are the same. 

(V) By hypothesis d(s'm x) = - dx .*. cos x = — .-. x = 6o°. 

/ 7\ 7/ • \ or V3 j d(s'mx) V3 

{d) (/(sin x) = cos 30 tfx = dx .-. — ^— — - = — - . 



23. Which increases more rapidly, an arc or its tangent ? At 
what value of the arc are the rates the same ? At what value 
of the arc is its rate \ that of the tangent ? At what value is 
its rate J that of the tangent ? 

24. Assuming that the ratio of the rates of change of an arc 
and its cosine remains constant while the arc changes from 
62 42' to 62 42' 25" find the natural cosine of the latter arc, 
given the natural cosine of the former as .45865. 

Ans. .45854. 

25. A fly-wheel connected with a stationary engine is revolv- 
ing uniformly at the rate of 2 turns a second. Compare the 
velocity of a point 1 foot from the axis with its horizontal 
velocity. 

Dropping a perpendicular from the point in any position to 
the horizontal line through the axis, we readily see that the 
horizontal velocity of the point is the same thing as the rate of 
change of cos x, where x — arc already described, estimated 
from the origin of arcs. Hence 

d (cos x) = — sin xdx , 

i.e., the horizontal velocity is sin x times the velocity in its path. 
Since dx = 2 tt .2. = 4 ir. 



68 Differential Calculus 

d (cos x) = — 4 7r sin x ; 
x = o, then d (cos .#) = o ; 

# = 30, then d(cos x)= — 2 7rft. a second = -J- velocity of point ; 
x = 90 , then d (cos #) = — 4 7r ft. a second = velocity of point ; 
x = 1 50 , then d (cos .%■) = — 2 7rft. a second = -g- velocity of point; 
x = 270 , then d(ccisx) = 4 7rft. a second = velocity of point. 

26. The crank of a steam engine is one foot in length and 
the coupling-rod is 6 feet ; find the velocity of the piston per 
second when the crank revolves uniformly at the rate of 5 turns 
per second. 




Let the length of crank = m and length of coupling-rod 
= 11. Let <f> = varying angle described by the crank and 
y = varying distance, the rate of change of which equals the 
velocity of the piston. 
Then n 2 = m 2 -\- y 2 — 2 my cos <£ ; 

.". jp 2 — 2 my cos cfi -|- m 2 cos 2 cf> = n 2 — m 2 -\- m 2 cos 2 <f> ; 

.-. y == m cos </> + V/2 2 — m 2 sin 2 </>. 

/ . , m 2 sin 2 d> \ 

Hence dy = - »ism<£ H Uf>. 

V 2 Nrr — nrswr^l 

I . sin 2 <£ 

But ;/z= i,« = 6, d§—\OTr.\ dy= — sm<£ H 7 — - . — )i0 7r. 



When <£ = o°, 
When <£ = 45 , 



2 V36 — sin 2 </> 



//y = o. 



dy = — [—=. H ) 10 7T 

W2 2 V36 -£/ 

— ( * H — 7= ) 5tt V2 feet per second. 



When (fj = go°, 
When <f> = 270 , 



Analytical Applications 

dy = — 1 o 7r feet a second. 
dy = 10 tv feet a second. 



69 



27. Compare the velocity of a train moving along a horizon- 
tal tangent with the velocity of a point at the base of the flange 
of one of the wheels. Compare also the vertical and horizontal 
components of the velocity of the flange point. 

The velocity of the train is the same as the velocity of the 
center of one of the axles. Call this velocity v. The point on 
the flange of the wheel describes a cycloid whose equations are 

x = ad — a sin 0, 
y = a — a cos 0, 

("Analytical Geometry," p. 206) in which a = radius of wheel 
and 




o b 

Fig. 9. 

= angle through which the wheel has rolled at the instant of 
consideration. 

Let jP, Fig. 9, be the position of the flange point at the in- 
stant ; then PCB = 0. We are to compare the velocities of P 
and C. 

Differentiating the equations of the curve, we have 

dx = a (1 — cos 6) dO, 
dy — a sin OdO. 

Since C is always vertically over the point B the velocity of 
C = rate of change of distance OB ; but OB = PB = aO ; 
.*. v = adO. 



70 Differential Calculus 

Hence, dx = v (i — cos 6), 

dy = v sin 6. 

From § 1 8, (3), we have 
ds= ^ldx* + df. 

Hence, ds = Vz' 2 (1 — cos 0) 2 + £' 2 sin 2 = z; V2 (1 — cos 0) 

. 6> 

= 2? sin — . 
2 

Now at O, = o° .'. dx = o, dy = o, ds = o, i.e., at the in- 
stant the point touches the rail it is at rest. 

At A, = 90 .*. dx = v, dy = v, ds == v V2, i.e., when the 
point is in a horizontal line with the center C both the hori- 
zontal and vertical components of its velocity are equal to the 
velocity of C, i.e., of the train, while the velocity in its path is 
equal to the square root of the sum of the squares of its compo- 
nent velocities. At D, 6 = 180 .*. dx = 2 v, dy = o, ds = 2 v, 
i.e., the vertical component of its velocity is zero while its hori- 
zontal component = velocity in its path is twice the velocity of 
the train. 

. 2 av . 

Again : since ds = 2 v sin - = sin — , we have 

2 a 2 

ds 2 a . 
— = — sin — . 
v a 2 

Q 

But 2 a sin - = chord PB and a = BC\ 

2 

ds chord PB 



v BC ' 

i.e., the velocities of P and C are proportional to the distances 
of the points from B ; hence B is the instantaneous axis about 
which the wheel revolves. 



Analytical Applications 71 

28. Assuming the rectangular equation of the cycloid, deduce 
the same results as in Ex. 27. 

y 1 

The equation is x = a vers -1 V2 ay — y 2 . 

See "Analytic Geometry," p. 207. Hence 

ydy 



dx 



v; 



' 2 ay — y 

y 
Since PB = a vers -1 - , we have 



a 



d(PB) = da vers -1 ^ = ady = v ; 

a V2 ay — y 2 



dy = - V2 ay — y 2 . 



Hence, dx = - v. 



a 

2 y 




Hence, ds = v 

t a 

At O, y = o .'. dy = o, dx = 0, ds = o. 

At A, y = a .*. dy = v, dx = v, ds = v V2. 
At Z), y = 2 a .'. dy = o, dx = 2 v, ds = 2 v. 

ds \l2ay chord PB 



Also 



v a BC 



Note. — The cycloid enjoys the mechanical properties of being the 
curve of quickest descent and of equal times. The problem of determining 
the line of quickest descent under gravity was proposed by John Bernouilli 
in 1696. The origin of the calculus of variations may be traced to this 
problem. Pascal applied the Method of Indivisibles of Cavalieri with emi- 
nent success to the investigation of the properties of the cycloid- 



72 Differential Calculus 



CHAPTER VI. 

GEOMETRIC APPLICATION. 

CARTESIAN CURVES. 

69. Tangent. The equation of a line passing through the 
point (V, y') is ("Analytical Geometry," p. 38), 

y — y r = s (x — x r ) 

in which s represents the slope of the line. We have seen 

dy 
§ (19) that — (= tan a) as derived from the equation, jy =/(x), 

represents the slope of the tangent to the curve at the point 

(x, y). If, therefore, we let -=—, represent the value of — at 
the point (V ', y f ), we have 

y-y' = %b-*) (*) 

for the general equation of the tangent to any plane curve. 



70. Normal. Since the normal at (x', y r ) is perpendicular to 
the tangent at that point, its slope is minus the reciprocal of 
that of the tangent. Hence 

dx 

y - y = - jj (* - * ) ( 2 ) 

is the general equation of the normal to any plane curve. 



Geometric Application 73 



EXAMPLES. 

1. Find the equation of the tangent and normal to the circle 

x 2 -j- y 2 = a 2 . 

„.~ . . dy x dy' x 
Differentiating, — - = .-. -£-= = 7 ; 

dx y dx y 

x r 
hence, y — y r = 7 (x — x r ) ; 

, . xx' yy' 

or, reducing, — + <~ = i 

(Is €& 

is the equation of the tangent, and 

y 

j - y = ^ o - x ), 

or, - = — 

y x 

is the equation of the normal. 



2. Find the equations of the tangent and normal to the 
parabola y 2 = 2px. 

dy p dy' p 

dx y ' dx r y r 

p 
Hence y — y r = —, (x — x'), 

or, yy' = p (x + x r ) 

is the equation of the tangent. 

dy' 
Substituting the value of —, in equa. (2) § 70, we have 

y 

y-/_ = -- (x-cc) 
for the equation of the normal. 



74 Differential Calculus 

3. Find the equations of the tangent and normal to the ellipse 

x 2 y 2 
a 2 b 2 

xx' yy r a 2 / . ,. 

a 2 cr ' b 2 x ' 

4. Find the equations of the tangent and normal to the 

hyperbola _ _ _ = I# 

xx' yy' , a 2 / ,. 



5. Given the equation 3 a 2 y = x 3 — 3 ax 2 + b\ find (a) the 
direction of the curve at the points whose abscissas are x = o 
and x = a and (b) the abscissas of the points where the curve 
is parallel to the ^c-axis. 

(a) The direction of a curve at any point is that of its tan- 
gent at that point ; hence, differentiating the equation, we have 

dy x 2 — 1 ax 

dx a 2 

dy 
At x = o, — = o .*. the tangent is || to the ^-axis. 
dx 

At x = a, — = — 1 .*. the tangent makes an angle of 135 
with x-2ixis. 

(b) Equating the value of — to zero, we have 



hence, 



are the abscissas of the points where the curve is || to x-2ods. 





X 2 


— 2 

a 2 


ax 


: 0; 




x(x 


— 2 


a) = 


0, 


.'. X 


= 


and 


x = 


2 a 



Geometric Application 75 

6. Does the direction of the curve in Ex. 5 at any of its 
points make an angle of 45 with the .x-axis ? 

Yes, at x = a (1 + V2) and at x = a (1 — V2). 

7. Find the equation of the tangent to the semi-cubic para- 
bola y% = aix at the point whose abscissa is a. 

Ans. 3 y — 2 x = a. 

8. Find the equation of the tangent to the cissoid ^ 2 = 



2 a — x 

, ?. ax 2 — x ? . .. 

Ans.y-y = ± (2 „ _ yt) (* - * X 

9. What angles do the cissoid y 2 = and the circle 

2 a — x 

x 2 -\- y 2 — 8 ax = o make with each other at their points of 
intersection ? 

Ans. At one point, 90 ; at two others, 45 . 



10. At what angle does the cissoid cut its base circle ? 

Ans. tan - 1 2. 

11. What is the equation of the tangent to x 2 (x + y) 
= a 2 (x — y) at the origin ? Ans. y = x. 

12. Show that all curves represented by the equation 

— +T = 2 

touch each other at the point (a, b). 

71. Length of Subtangent. Length of Tangent. 

Let FT, Fig. 10, be the tangent to MS (y =/(x)) at the point 
P{x\y'). 



76 



Differential Calculus 



PA / 

Then, Subtangent =^r=/^ cot ATP= -J 7 ri> = ~T' '■> § J 9- 

tan ^l j. Jr ay 

dx 1 




Fig. io. 



.'. Subtangent = y 



, dx f 
~dj/' 



Again, Tangent = PT= ^ AP 2 + A T 2 = ^ y' 2 + y' 2 (^-)j ; 



.-. Tangent =/y i +^j. 



72. Length of Subnormal. Length of Normal. Perpendicular 
to Tangent. 

Let PB be the normal to the curve at P, then 

Subnormal =AB =AP tan APB =APt<mATP. 

dy 
.-. Subnormal = / ——. . 
dx 

Again, Normal =PB = \J~AP 2 + AIP= \J/ 2 +/ 2 (j~) , 

,. Normal =/ V / I+ (gJ i 

Draw 0C1 PT, then 



Perpendicular = OC= 



or 



OA-AT 



x — y 



,dx r 

w 



csc ATP ^ 1+ cot *A TP k I (dx'V 

+ W1 



v/- 



.-. Perpendicular = 



x r dy r —y r dx r 
(dy' 2 + dx' 2 )^ 



Geometric Application 77 



EXAMPLES. 

1. Find the lengths of the subtangents and the subnormals 
of the conic sections. 

Circle. Parabola. Ellipse. Hyperbola. 

a 2 -x' 2 . x' 2 -a 2 x' 2 -a 2 

Subtangents -, — 2 x 

Subnormals x r p 



b 2 x tfx' 

a 2 a 2 



As lengths only were required the signs of the values are 
omitted. If we take Z"and A, Fig. 10, as points of reference, 
the signs will show whether the subtangents and subnormals are 
measured to the right or to the left. 

2. Find the length of the normal to the catenary y = — (e a -{-e «). 

y * 

Ans. — . 
a 

3. Deduce the equation of the tangent to the hypocycloid of 
four cusps, x /s -\-y /,s = c/ s , and show that the position of the tangent 
included between the coordinate axes is constant and equal to 
the radius of the base circle. 

4. Find the lengths of the subtangent and subnormal to the 

. . _ x 3 (2 a — x'~)x' (^a — x)x' 2 

cissoid jr= . Ans. 7— , — j-r . 

2 a — x 2> a ~ x {2 a— xy 

5. Show that the subtangent of the hyperbola xy = m is 
equal to the abscissa of the point of tangency. 

X 

6. Show that the subtangent of the logarithmic curve y = ae c 
is constant and equal to c. 

7. Show that the values of the normal and subnormal of the 

cycloid, x = a vers -1 V2 ay— y 2 , are V2 ay' and \J(2 a—y')y f , 

respectively, and from these values show that the line joining 



78 Differential Calculus 

the generating point and the foot of the vertical diameter of the 
rolling circle is always normal to the curve. See Ex. 27, p. 69. 

8. Find the length of the perpendicular let fall from the 
origin to the tangent of the hypocycloid sr-\~y *= a 3 . 

Ans. yax'y'. 

73. Rectilinear Asymptote. Equations of the Asymptote. 

The limiting position of a tangent to a curve as the point of 
tangency recedes to an infinite distance is called the rectilinear 
asymptote of the curve. 

Of course curves with infinite branches only can have 
asymptotes. 

Assuming the general equation of the tangent to any plane 

dy 
curve, § 69, y — y= —, (x — x'), 

and making successively y = o, and x= o, we obtain 

dx 

T *= x -y „y> 



dx 



/, = /-*'^7, 



for the intercepts of the tangent on the X-axis and the K-axis. 
Now, if either of these intercepts approach a finite limit as 
either coordinate, x r or y r , of the point of tangency approaches 
an infinite value there is an asymptote whose equation may be 
determined in either of the following ways : 

1., By ascertaining the limits of I x and I y , i.e., by determining 
the two points in which the asymptote cuts the axes ; or, 

2., By ascertaining the limit of one of the variable intercepts 

dy' 
of the tangent and the limit of its slope, -~y, as the tangent 

point recedes infinitely. 



Geometric Application 79 



It frequently happens in the effort to ascertain the limit of 

dy' 
Z., or L„ or -r-p that the values assume an indeterminate form. 
y dot 

If so the process of evaluation is determined by principles ex- 
plained in Ch. IX. 



First Method. By ascertainifig the limits of I x and I y . 

Let x / and y J be the limits of I x and I y as x' or y f approaches 
an infinite value ; then 

- + -=i (i) 

is the equation of the asymptote in its symmetrical form. 

dy r 
Second Method. By ascertaini?ig x, or y , and the limit of -^—. . 

' - 71 J dx r 



(a) If we determine y / and Limit 



dx r 



X': 



, we have 



'df 
dx' 



y = Limit 
for the slope equation of the asymptote. 



x+yj (2) 

y'= 00 



r~//i/ 

(F) If we determine x t and Limit ~— f 



j we have 

V'= «> 



%\y + x < (3) 

for the equation of the asymptote. 

Cor. 1. If Xj = o necessarily y / = o, and if y — o necessarily 
x t = o ; i.e., the asymptote passes through the origin. 

Cor. 2. If x / = 00 and y j = 00 there is no asymptote. 

Cor. 3. If x / = 00 and y j = b = some finite quantity, then, 
equa. (1) becomes 7 = b, 



80 



Differential Calculus 



or, if 



Limit 



dx' 



x' = 



o and_y / = b, th,em equa. (2) becomes 



V'= °° 



y = b, 
that is, in either of these cases the asympfete is || to the jc-axis. 
Cor. 4. If y J = 00 and 'x = a, then equa. (1) becomes 

x = a ; " 



or, if Limit 



dx 
ly'_ 



o and x = a, then eo^ua. (3) becomes 

x = a. 



That is, in either of these cases the asymptote is || to the 
Kaxis. As the second of the two methods explained above 
for determining an asymptote is usually the simplest we shall 
adopt it in the following. 



EXAMPLES. 

X V 

1. Examine the hyperbola — — — = 1 for asymptotes. 



Here 



dy' _ b\x b I £ 2V 

dx' d 2 y ~ a \ y" 2. 



.*. Limit 


\dx'_ 


x'= 00 

y'= °° 


= Limit 


[*:( 




y'= 


b 
= ± — 
a 


= Limit 




= Limit 

y'= °° 


ay 


y' = 00 


= Limit 


' IF 

_ 7. 


yi— 00 


= 0. 











Substituting these values in equa. (2) § 73, we have, 



v = ± — x 

y a 



for the equations of the asymptotes. 



Geometric Application 
2. Examine the parabola^ 2 = 2 fix for asymptotes. 

Here 

dx 

dy 



81 



dy' _fi m 

dx' y 



.'. Limit 



dx 



= o 



y'= 00 



.-. if there is an asymptote it is || to the X-axis ; Cor. 3, § 73. 

X r = 00 



y = Limit 



y — x' 






= Limit 



00 . 



_ y'= 00 



Therefore there is no asymptote to the parabola. 

3. Examine the curve y 3 — x 3 -\- a 2 x for asymptotes. 

The equation solved for y shows that as x = 00 , y = 00 and 
as x = — 00 , y = — 00 ; /. the curve has infinite branches in 
the first and third angle. 

,2 



dy x 2 

Here — = — -\ 5 

dx y 2>T 



V XT 



+ 



3 \l x 2 (x 2 + ^ 2 ) 2 ' 



•. Limit 



■df 

dx' 



= Limit 



X 1 = i 00 



(1 + ^) 3V^+^ 



= 1. 



-1 a/= ± qo 



.*. the asymptote makes an angle of 45 with ar-axis. 



jy == Limit 



y —x 



X f = 4- 00 



dx' 



= Limit 



id 1 



X'i+— X 

X ?/ -> ~/—±, 



o. 



X' = ± 00 



.-. Equa. (2), § 73, y = x 

is the equation of the asymptote. 

4. Examine y B = x s -{- ax 2 for asymptotes. 



y4«J". y — x = 



82 Differential Calculus 

74. Asymptotes by Inspection. The limiting position of the 
tangent to a curve as the point of tangency recedes infinitely is 
evidently a straight line which the curve continually approaches 
but never reaches. Taking this view of a rectilinear asymptote 
we are frequently able to determine the equation of the asymp- 
tote by simply inspecting the equation of the curve. 

EXAMPLES. 

x 3 

1. Determine the asymptote of the cissoid v 2 = 

2 a — x 

We see from the equation that as x approaches the value 2 a, 
y approaches an infinite value, .*. the curve continually ap- 
proaches but never reaches the line x = 2 a. 

.'. x = 2 a 
is an asymptote. 

4. fPx 

2. Examine the witch y 2 = for asymptotes. 

2 a — x 

We see that x = 2 a is an asymptote and as the curve has 
infinite branches in the first and fourth angles only there is no 
other asymptote. 

3. Examine the conchoid x 2 y 2 = (J? — y 2 ) (a + y) 2 for asymp- 
totes. 

Here x = ± °-^- V/> 2 - y 2 . 

y 

Asjy = o, x = ± 00 , .*. y = o, i.e., the ^r-axis is an asymptote. 

4. Examine the curve of tangents for asymptotes. 

TV 3 7T 

We see from the equation y = tan x that as x = — , or - — , or 

2 2 

- — , or etc., that y = 00 , or — 00 ; .'. 
2 

t 1 3 * i 5 7r ■ 

x = - and x = - — and x = — , etc., 

222 

are asymptotes. Similarly we may show that 



Geometric Application 83 

7T 3 7T 5 7T 
X — , X — , X — ^ cLC.j 

2 2 2 

are also asymptotes to the curve. 

5. Examine (x — 2 a) y 2 = x s — a s for asymptotes. 

By inspection we find x = 2 a is an asymptote. 

By analysis we find two others y = x -f- a and y -\- x -{- a = o. 

75. Asymptotes by expansion. Where an equation can be 
readily solved for one of the variables and the second member 
can be expanded into a series the asymptote may frequently be 
more readily detected than by pursuing the general course as 
explained in § 73. 

EXAMPLES. 

1. To find by expansion the asymptotes of the hyperbola. 
9, *> 

x V" 1 

From the equation — — = 1, we obtain 

a z b 2. 



bx 


:- 


a 2 \i 
x 2 / 


bx 

= + — 


(■- 


1 a 2 


a 


\ 


2 x 2 



Sx 4 16 x & 

As x increases indefinitely, the curve approaches nearer and 
nearer the lines represented by the equations 

bx 

y= ±— ; 

a 
.'. these equations represent the asymptotes to the curve. 

2. To find by expansion the asymptote of x* — xy 2 + ay 2 = o. 

/ x s \* / 

Here y = ± [ = ± x 1 — 

\x — a \ x 



84 Differential Calculus 

/ a 3 a 2 5 a 3 

\ 2 x 8 x 2 i6jc 3 

a' 

.-. y = ± \x + - 

are equations of asymptotes. By inspectio?i we see also that 
x = a is the equation of an asymptote. 

3. Find by expansion the asymptotes of the curve represented 
by the equation y 2 — 2 xy — x 2 + 2 = o. 

Ans. y = (1 ± A ^2)x. 

4. Find by extraction of the root the asymptotes of the 
curve y 4 — 96 a 2 }? +100 cPx 2 — x 4 = o. 

Here y = ± \ 48 <? 2 ± V2304 # 4 — 100 a 2 ^ + x 4 

/ a 2 

= ±[x 

\ x 



99 a * 

2 x 3 



.'. y = ± x 
are the equations of the asymptotes. 

POLAR CURVES. 

76. To find the slope of the tangent to a polar curve. 

From § 19, we have 

dy 

tan a = — 
dx 

for the slope of a curve when referred to rectangular coordinates. 
If we assume the pole coincident with the origin and the initial 
line coincident with the jc-axis we have, " Analytical Geometry,' 

Art. 34, (3), 

x = r cos 0, y = r sin 0. 



Hence, 



Geometric Application 85 

dy d (r sin B) r cos 6 d6 -\- sin 6 dr 



dx d (r cos 6) cos 6 dr — r sin $ dO 
is the required expression. 

Cor. from § 18 (3), we have 

ds = ^dx 2 + dy 2 ; 
.-. ds = V '\d \r cos 6)\ 2 + {d(r sin 0)\ 2 = \Jdr 2 + r 2 dd 2 . 

77. Length of Subtangent. Length of Tangent. 

Let MS be any curve referred 
to O as pole and OX as initial 
line. Let P (r, 6) be any point 
of the curve at which a tangent 
PB and a normal PA are drawn ; 
then OB and 6X4, segments 
of the perpendicular to the 
radius vector OP, are respec- 
tively the subtangent and sub- 
normal corresponding to the 
point P (r, 6). 

1. To find a value for OB, the subtangent. 
From the triangle OPB 

OB = r tan <f> ; but 




tan <f> = tan (a — 6) = 



tan a — tan 



1 + tan a tan 6 

r cos dO + sin dr sin 
cos dr — r sin d# cos 



1 + 



r cos dO + sin dr sin 



Art. (76), 



tan <£ = 



cos dr 
rdO 



r sin dO " cos 



dr 



Subtangent = r 



d0 
dr 



86 Differential Calculus 

2. To find a value for PB, the taiige7it. 
From the triangle OPB 



PB = V? + QB 2 =\J r 2 + r 2 ~ ; 



1 +1? 



78. Length of Subnormal. Length of Normal. Perpendicular 
to Tangent. 

From triangle OAF, Fig. 1 1 , 

oa op r r 



tan OAF tan </> dO' 

dr 



dr 
Subnormal = -y- ; 
do 



also, PA = ^OP 2 + OA" = V^ 2 + 



dr 



.-. Normal = yV+-^. 

From triangle OCB (OC being perpendicular to BP), we 
have 

_ OB dr r dr 



sec COB Vi + tan 2 <j> 4 / r 2 d$ 2 



dr 

.-. Perpendicular = 

'dr 2 2 

h r z 

dO 2 



Geometric Application 87 



EXAMPLES. 

1. A circle whose diameter is a is referred to the lower ex- 
tremity of its vertical diameter as a pole, and to the tangent at 
that point as an initial line. Find the relation between <f>, a, 
and 0. 

The equation of the curve is evidently r = a sin 0, 

d& r a sin . 

:. tan <£ = r — = — = = tan 6 ; 

dr dr a cos v 

Je 

Fig. ii, a=<f>-\-6=20=2<f>. 

2. Show that the logarithmic spiral r = a e is an equiangular 
spiral, i.e., that the tangent makes a constant angle with the 
radius vector. 

r a 9 i 

tan <p = 



dr a e log a log a 
TO 

If a = e, then tan <f> = i, .*. <f> = 45 . 

3. Show that the perpendicular from the pole to the tangent 
of the lemniscate r 2 = a 2 cos 2 $ varies with r 3 . 



v/ 



2 dr 2 \ict (cos 2 2 + sin 2 2 0) 



17 



Show also that <£ = — \- 2 6 . 



2 



4. In the spiral of Archimedes r = cB show that the product 
of the subnormal and subtangent is always equal to the square 



88 



Differential Calculus 



of the radius vector. Find also the value of the tangent, 
normal and perpendicular. 



Subtangent = t* — = — 
dr c 



Subnormal = 



dr 

Jo 



.'. Subtangent X Subnormal = r 2 . 
Tangent = - *s/c 2 + i. 



Normal = \[^ 
Perpendicular = 



+ c\ 



\/r 2 + c 2 

5. In the hyperbolic spiral rO = c show that the polar sub- 
tangent is constant and equal to the circumference of the 
measuring circle. 

6. Show that the area of the square formed by tangents to 
the cardioid r = a (i + cos 6) inclined at an angle of 45 to 
the initial line is f £ (2 + V3) a 2 . 

79. Asymptotes. 

Let r =/(&) be the equation of MS, Fig. 12, and \ttAB be 

an asymptote. Draw OC || to AB and 
OB _LOC; then OB is _L to AB. Now 
OC = r is the radius vector of the in- 
finite tangent point, and OB is the 
subtangent corresponding to that point. 
If, therefore, 6' be the limiting value of 
6 as r approaches 00 as its limit, we 
have 




and 



Limit [f(0y\o=o'= & , 
~r 2 d6 



Limit 



dr 



= OB = a finite value, 



6=0' 



as the conditions for an asymptote to a polar curve. 



Geometric Application 



89 



Looking in the direction of the infinite radius vector OC, the 
distance 



OB = limit 



dr 



0=0' 



is laid off to the right or left according as OB is positive or 
negative. 

EXAMPLES. 

1. Examine the curve of tangents r — c tan for asymptotes. 
Here, Limit [f(O)~\ e i ,= Limit |Vtan 0] fl _L + 5 = oo ; 

~ 2 

also, Limit r 2 — - = Limit \c 2 sin 2 &]■■* = c. 

L ar\ e=e , J0 =~ 2 




Hence drawing the curve and the radii vecrores correspond- 

ing to the vectorial angles - and , we see that BA J- to OX 

at a distance C from O and B' A' J_ to OX and at the same 
distance C from O are asymptotes to the curve. 



90 



Differential Calculus 



Here 
and 



2. Examine the lituus r 2 6 = c for asymptotes. 



Limit [/(ff)]o=e'= Limit 
d9 



= 00 



Limit 



dr 



6=6' 



= Limit [— 2 r6] e=0 = o. 



Hence, the initial line is an asymptote. 

3. Examine the hyperbolic spiral rO = c for asymptotes. 



Limit \_f(Qy\e = e>— Limit 

M 



= oo 



»=0 



Limit 



dr 



6=6' 



Limit [— c] e / = o= — c\ 



.'. a line || to the initial line and at a distance c above it is an 
asymptote. 

4. Taking the left-hand focus as a pole, examine the hyper- 
bola for asymptotes. See "Analytical Geometry," Art. no, 
Equa. 3. 

5. Examine the following curves for asymptotes, 
r = 2 a tan $ sin 6. r 2 cos = a 2 sin 3 0. 

r = a sec 6 ± b. rcos 2 $ = a (1 -f- sin 2 0). 



Successive Differentiation 91 



CHAPTER VII. 

SUCCESSIVE DIFFERENTIATION. 

80. Successive Differential Equations. — If we differentiate 
the expression y = x^ we obtain 

dy = 4 x z dx. 

Since dy is a function of # it can be differentiated. Regarding 

x as equicrescent, i.e., as changing uniformly, we have dx = a 

constant. Hence differentiating again, and representing d(dy) 

by d 2 y, we have 

d 2 y = 12 x 2 dx 2 , 

in which d'x 2 = (^r) 2 . Differentiating again, and representing 

d(d 2 y) by d s y, we have 

d s y = 2\xdx z , 

in which ^r 3 = (^c) 3 . The differentiation may obviously be 
continued until the second member vanishes. 
Hence, in general, if 

y =f(x), we have 

dy=f r (x)dx (a) 

d 2 y =/" (*) dx 2 (b) 

d % y=f' r, {x)dx z (c) 

d 4 y =/ iv (*) dk 4 (d) 

d n y =f n (x) dx" ....... . (V) 

in which (#), (#), (V), (//) ...(e) are called respectively the 
First, the Second, the Third, the Fourth . . . the nth Dif- 
ferential Equation of the equation^ =f(x). 



92 Differential Calculus 

81. Successive Differential Coefficients or Derivatives. 

If we divide equations (a), (&), (V), (d), . . . . (e) by dx, dx 2 , 
dx 3 , dx*, . . . dx 11 , respectively, we have 

!=/'(*) ^ 

d 2 v 

a?--™ <"> 

3=^) <*> 

These equations are called respectively, The First, The Sec- 
ond, the Third, the Fourth, . . . the nth Differential Coefficient 
or Derivative of the equation y =f(x). 

Cor. As f n (x) is the first derivative of f n ~ x (x) it follows 
§ (28) Cor. that f n (x) is positive or negative according as 
f n ~ x (x) is an increasing or decreasing function of x. 

EXAMPLES. 

1. Write the successive differential equations and coefficients 

of y = x 5 , 

dy = c x 4 dx .'. -r- = K x*. 
dx ° 

d 2 y 
d 2 y = 20 X s dx 2 .'. —r~ = 20 x 3 . 

dx 1 

d 3 y 
d 3 y = 60 x?dx 3 .*. -7-5 = 60 x 2 . 

dx 3 



7 . . a y 

dy = 120 xdx .'. ——, = 120 x. 



d*y 
dx* 



Successive Differentiation 93 

d 5 y = 120 dx 5 .'. -r-z = 120. 
J dx> 

d e y = o .-. —4 = o. 

J dx b 

2. Write the successive differentials of y = 2 x 3 — 3 x 2 + 7 x. 
dy = (6 x 2 — 6 x -f- 7) dfc, 
^/ 2 j = (12 .r — 6) dx 2 , 
d s y = 12 dx 3 , 
d 4 y = o. 

Find the differential coefficient indicated by the answers in 
the following : 

1 — x 

4. jy = x i log x. 

5 . y = tan # + sec x. 

6. j = e~ x cos jf. 

7. j; = (x 2 + # 2 ) tan -1 - 

8. y = log (sin x). 

9. ^+y = a 2 . 

10. a 2 y 2 + ^ 2 = tf 2 ^. 

11. ^ = 2^JC. 

12. a 2 y 2 - Px 2 = - a 2 £ 2 . 



A' 



fl! 



dx 4 


(' 


- xf 


dy 




\± m 


dx 6 ~ 




X 2 


d 2 y 




COS Jt: 


dx^~ 


(1 


— sin x) 2 


d 4 y 
dx 4 




4 ^ _a: cos X. 


d 3 y 




4 a 3 


dx 3 


(a 


2 + xf' 


d 3 y 


2 


COS X 


dx 3 


sin 3 jc 


d 2 y 




9 


dx 2 




7' 


d 2 y 




^ 4 


dx 2 




a 2 y 3 


d 2 y 




f 


dx 2 




/' 


d 2 y _ 




* 4 


dx 2 




« 2 _>> s 



94 Differential Calculus 

By examining the successive derivatives we can frequently 
express the n xh derivative of a function. 

d n v 

13. y = a x . —^ = (log dfa x . 
J dx 11 v te } 

d"y 

14. y = /' r . -/- = n n e ax . 
J dx n 

d n \n- I (- I)"" 1 

15. y = log x. 



16. y 



dx n x n 

i —x d n y __ 2 LZ£ C — 0" 

i -+- x ' dx* 1 ~ (i + x) u + 1 



APPLICATIONS. 

82. Definition. Acceleration is the ?-ate of change of velocity. 
Let a = acceleration and v = velocity ; then 

dv d 2 s , . ds 

a = di = dF* (***» = #• §I ?> 

1. A body, originally at rest, falls in a vacuum near the 
earth ; find its velocity at any instant, and the acceleration of 
that velocity. 

From Mechanics we have 

s — 2 & l 

for the distance s fallen by a body in the time /. 

Differentiating successively we have, if we do not regard 

dt = unit of time, 

ds 
ds = gtdt .'. v = — - = gt ; 
6 dt * ' 

d 2 s 

Hence the velocity of a falling body is a variable, while the 
acceleration of that velocity is constant. 



Successive Differentiation 95 

2. A projectile thrown obliquely upward at an angle de- 

g 

scribes a parabola whose equation is y = x tan — 



2tr i COS 2 

Assuming the initial velocity (= v), and the horizontal compo- 
nent of this velocity (dx = v cos = a constant), find the 
velocity of the projectile in its path, and the acceleration of its 
velocity vertically. 

Here 

g 

dy == tan 6 dx = 7r -xdx, 

ir cos- 

= v sin 6 — j. x = vertical component. 

V COS V 

d 2 y = — g = acceleration of velocity vertically. 
ds = 's/dx 2 + dy 2 = "vV — 2 gy = velocity in its path. 

3. The distance described by a point whose initial velocity 
was u is given by the equation s = ut + \ ct z ; find its velocity 
at any instant and the acceleration of the velocity. 

Am. v = u + \ ct 2 , a = ct. 

4. The generating point of the parabola y 2, = 2px moves 
with a constant velocity v' ; find the velocities and accelerations 
in the direction of the axes. 

P 
Here dy = - dx. 

y 

Since 7/ = ds = ^jdy 1 -f- dx 2 = dx, 

y 
y .... 

we have dx = , v = velocity in direction of x ; 

P 
hence, dy = , if = velocity in direction of y. 



96 Differential Calculus 

Differentiating these values, we obtain 

f 
d* x — / ^ , — ^ v'* — acceleration in direction of x. 

(f +r) 2 

d 2 y = — 7-75-; — ^ ^ /2 = acceleration in direction of y. 

Hence, since d 2 x is always positive, its function, the velocity in 
the direction of x, is an increasing function ; since d 2 y is posi- 
tive in the fourth angle and negative in the first, the velocity in 
the direction of y is increasing in the fourth and decreasing in 
the first angle. The student should bear in mind that the 
terms increase and decrease are used in an algebraic sense. 

5. A point describes a circle of radius r with uniform ve- 
locity v. Show that the resultant acceleration at any position 

in its path is — • 
r 

y 

6. The generating point of the cycloid, x = a vers -1 

V2 ay — y 2 , so moves that the component of its velocity in 
direction of x is constant and = m ; find the velocities and 
accelerations in the direction of y and in its path. 



Here 



y y 




j i/ 2a 

as = m \ — > 
V y 




d 2 y = - y >» 2 , 




in 2 la 
d 2 s = - y-( 2 a-y). 





83. Theorem of Leibnitz.* To find an expression for the nth. 
differential coefficient of the product of two variables which are 
functions of a third variable. 

* Leibnitz published this theorem in 1710. 



Successive Differentiation 97 

Let u and v be functions of x, and let u 1} u 2 ,u z , etc., and v lf 
v 2 , v s , etc., represent the successive derivatives of u and v. Let 
y = uv\ then, § 25, 3, 

dy 

—— = uv x -\- vu x . 

ax 

d 2 y 
Hence, -7-^ = uv 2 -\- tc 1 v 1 -f- zy'i + z^ 2 = ^2 + 2 *Vi + ^2- 

</ 3 y 
Hence, — z = uv s + u x v 2 + 2 (//^ + v t u 2 ) + zy/ 2 + zw 8 

= «z> 3 4- 3 z^ -f 3 k^ + zw 3 . 

We observe that the coefficients in the second derivative are 
the same as those in the expansion (v -f- uf = v 2 -f- 2 uv -j- u 2 , and 
that those in the third derivative are the same as in the expan- 
sion (v + iif = z' 3 + 3 uv 2 + 3 u 2 v + z* 3 . We further observe 
that the subscripts in the derivatives are the same as the expo- 
nents in the expansions, except that u enters the first term and 
v enters the last term. It appears, therefore, that the values of 
the successive derivations follow the laws of the Binomial 
Theorem both as regards the coefficients and subscripts, except 
in so far as indicated as to the subscripts. Assuming that the 
law holds for the nth. derivative, we have 

d n y n (n— 1) 

~^ = uv n +nu 1 v n _ 1 -{ ^ ^ 2 7/ ft _ 2 -r- etc., . . . 

-f nu n _ 1 v 1 + vu n ..."(«) 

Differentiating again and collecting, we have 
d n + 1 y , . (« + i)« 

-fap + i = UV n + X + \ 71 + *) U l J n-\ V~ 2 «2»«-l+ etC > ' ' • 

+ (n + 1) ^^ -f- vu n+1 ...(b) 

Hence the law still holds. If n is any integer formula (h) 
shows that the law holds for the next higher integer. But we 
have shown that the law holds for the integer 3 ; hence it holds 



98 Differential Calculus 

for the integer 4 ; hence it holds for the integer 5 , etc. Hence 
the formula (a) holds for all positive integers. 
In the usual notations formula (a) reads 

d n (uii) d n v du d n ~ 1 v n(ri— 1) d 2 u d n ~ 2 v 
^ L = u _j_ n _) ^ ' . 

dx n dx 71 dx dx 7l ~ 1 1 2 dx 2 dx n ~ 2 

d n ~ 1 u dv d n n 

+ etc., . . . 11— zr — -\-v — — 

dx' 1 ' 1 dx dx 11 

EXAMPLES. 

1. Find by Leibnitz Formula the third derivative of x 2 ^. 
Here v = <? ax , u = x 2 , 

dv _ du _ 

dx dx 

d 2 v _ „„ d 2 u 

= a 2 e ax = 2 

dx 2 ' dx 2 ' 

d * V 3 ax d *U 

— = a 6 ' . — = o. 

dx 3 dx 3 

.'. , ' = x 2 . a 3 <f* + x . 2 x . a 2 e ax -f ^~ . 2 . ae™ 

dx ° [2 

= ae ax (a 2 x 2 -f 6 ax -f- 6). 

2. Write the nth derivative in the example above. 
d n (x 2 ^ x ) . . n(n — 1) 

//•*" [2 

3. Write the /zth derivative of x 2 ^. 

d n (x 2 a x ) 



dx n 



(log #) re 2 a x \(?i -\- x log a-) 2 — /z \ 



4. y = x\ogx\ find ;zth derivative. 

d n y \ n - 2 , 

»n v m — 1 v / 



dx n x n 



Successive Differentiation 99 

5. y = £ -rE cos x; find 4th derivative. 

d*y 

-r—. = — 4 e x cos a;. 
ax* 

84. To find the values of the successive derivatives when neither 
variable is equicrescent. 

In the preceding articles and examples we have derived the 
successive derivatives under the supposition that x was equicres- 
cent, i.e., that dx was constant and therefore d 2 x = o, d s x = o, 
etc. If the variables are not equicrescent, then both dx and dy 
are variables. 

Let y =/(x) ; then 

is the first derivative whether x or y, or neither is equi- 
crescent. Differentiating again and remembering that — is a 
fraction with a variable numerator and denominator, we have, 

f dy 



dx) dxd 2 y — dyd 2 x 

-^r = i? — =/(x) •••■(») 

for the second derivative when neither x nor y is equicrescent. 
Differentiating again and collecting, we have 

'dy y 



d\ V x j 



dx J (d z ydx — d z xdy) dx — 3 (d 2 ydx — d 2 xdy) d 2 x 



dx dx b 

=/'"{*) (3) 

for the third derivative when neither x nor y is equicrescent ; 
and so we may continue the process until any desired derivative 
is reached. 

LofC. 



100 Differential Calculus 

Cor. i. If x is equicrescent, we have from the equation 
above, since d z x = o and d 2 x = o, 



^ 
^ 



.=/"'(*>■ 



Cor. 2. If y is equicrescent we have from the same equa- 
tions, since d 2 y = o and d z y = o, 

dx dx ^ 

dy 

d 2 x 



=_ 41 =/», 



d 2 xdy 
dx z /dx\ 3 



d 2 x\ 2 d s x dx 



7.(d 2 x) 2 dy — d s xdydx \dy 2 J dy s dy 

and -^^h — — = W =/"'(*)• 



dy) 

Scholium. It frequently becomes convenient in applying 
the principles of the calculus to change the equicrescent vari- 
able in a differential expression, thus converting the expression 
into an equivalent one under another form. 

Thus (1) if we desire to change a differential expression 
deduced under the assumption that x was equicrescent into an 
equivalent expression in which y is to be considered equicres- 
cent, we merely substitute the successive derivatives under 
Cor. 2 for those under Cor. i. 

If (2) we desire an equivalent expression in which neither x 
nor y is considered equicrescent we substitute for the succes- 



Successive Differentiation 



101 



sive derivatives under Cor. i the expressions (2), (3), etc., 
above. 

If (3) we desire to introduce a third variable z, a function of 
x or y, as the equicrescent variable we first proceed as under 
(2) and then substitute for the variables and their successive 
derivatives their values drawn from the given functional relation. 

EXAMPLES. 



1. If y = 
Here 



1 4- x 


dx 
show that — : 
dy 


\y 




1 — X 


<S + 


if 


dy 




I 






dx 





— x) Vi - 


- X 2 






- /t 










v^ \/t . 


~2 _ 


4_y 



" dy (f + i) 2 

Tr d 2 yf d 2 y /dy 2 \ dy d s y - , , d 3 x d 2 x 

2 - if ■&(*;& - (i) ) - dxdx->=°> show that z? + ^ 



= o. 



Replacing the derivatives by those given in § 84, Cor. 2, we 
have, 



d 2 x 
df_ 
dx x3< 3 
dy, 



d 2 x 
If 
~dxV 

dy) J 



r 



dxV dx 

dy) J dy 



/d 2 x\ 2 dx d 3 x 
3 \df) ~ dy df 
dx\ 5 
dy) 



= 0, 



/d 2 x\? d 2 x dx /d 2 xV dx d 3 x 



, + 



+ 



\dy 2 ) df dy *\df) dydf 



d s x d 2 x 
df df 



dx 



. d 2 y 2 x dy y 

dx 2 1 -\- x 2 dx (1 -f- x 2 ) 
-(- y = o where z = tan -1 x. 



d 2 y 



— o. Show that 

dz 



102 Differential Calculus 

Here we proceed as explained in (3), § (84), scholium. 

dxd 2 y — dyd 2 x 2 tan z dy y 

dx 3 1 + tan 2 £ dx (1 + tan 2 ^) 2 

Since x = tans, we have 

dx = sec 2 zdz and d 2 x = 2 sec 2 2 tan zdz 2 . 

sec 2 z(d 2 ydz — 2 dydz 2 tan z) 2 tan zdy y 
sec 6 zdz 3 sec 4 zdz sec*z 

d 2 y 2 tan zdy 2 tanzdy 
dz 2 dz dz 

d 2 y 



2 +y = o. 



dz 

4. If x 2 = 4 z, show that -z— 2 -\ — + y = o becomes 

5. If # = cos0, show that (1 — x 2 ) ■—, — x~ = o becomes 

v dx 2, dx 

d 2 y 

d0 2 = o - 

d 2 y x /dy\ 2 dy 

6. If x = jjtf 2 , show that x —4 I ^- ) + -r- = o becomes 

rt\3r jy \"^/ dx 

d 2 z dz 

v 1 = o. 

dy 2 dy 



7. If x = a cos 0, and jy = b sin 0, show that 
'dy" > * 



1 rtfr/ ( _ (« 2 sin 2 + ^ 2 cos 2 0)3 
dx 2 



Successive Differentiation 103 

8. Given x = r cos and y = r sin 0, find the equivalent of 



dx > 

— - — — = p, (i) when is equicrescent, (2) when r is 

equicrescent : 

I. When is equicresce?it : 

Replacing the derivatives by their equivalents in § 84, (1) 
and (2), we have 



\dxj 



dx z _ (dx 2 + ^2)1 

// 2 _y dxd 2 y — dyd 2 x dxd 2 y — dyd 2 x 

dx 2 dx 3 

for the general value of p when neither 3: nor y is equicrescent. 
Differentiating the values of x and 7, 6 being equicrescent, 

we have 

dx = cos Odr — r sin Odd, 

d 2 x = cos 6d 2 r — 2 sin OdOdr — r cos &Z0 2 , 
dy = r cos &/0 + sin ^^> 
^/ 2 j; = sin 6d 2 r -|- 2 cos &/&/r — r sin &/I9 2 . 

Hence, substituting and reducing, we have 



■ 2 U - r ^ 



II. When r is equicrescent. 

Differentiating the values under this supposition, we have 

d 2 x = — r sin 6d 2 6 — 2 sin OdrdO — r cos 0d(P 
d 2 y = r cos $d 2 + 2 cos OdrdO - r sinOdO 2 . 



104 Differential Calculus 

Substituting these values together with those of dx and dy 
deduced under I. as they are unaltered by the new supposition, 

we have 

<dfr 



~lFo ~le z de = p ' 

dr 2 dr 3 dr 



9. Show that when the equicrescent variable is changed 

dy" 



\ dx 

from x to y in p = -i — — - — — we have 

dy 

dx 2 

i/dxV 
I + W 

P = 



d> 2 



10. Change the equicrescent variable from x to y in the ex- 
d 2 z a 2 dz 
dx 2 x dx 



/ 9 9\ "" % & rf% ' 9 i 9 

pression {ar — xr) — - — z = o, given xr -f- jr = a 



A 2^ Z 

AllS. X* -r-i — z = o. 

dy 2 



Series 105 



CHAPTER VIII. 

SERIES. 

History. — Previous to the 17th century infinite series rarely occurred 
in mathematics. During the latter part of this century, and in the 18th, 
they came into very general use. It was supposed at this time that all 
higher calculations could be made to depend upon them. No universal 
criterion for determining the question of convergency or divergency was 
known at the time, nor, indeed, is one known to-day. James Gregory 
(1638—1675) was the first to draw a distinction between convergent and 
divergent series .; yet Cauchy and Abel, distinguished mathematicians of 
the 19th century, were the first to question results based upon them. New- 
ton's first mathematical discovery — the binomial theorem — was the first 
important contribution to this subject. The first rigorous proof of this 
theorem was given by Abel (1 802-1 829). Taylor's theorem, published in 
17 15, was the first general theorem on series published. The first correct 
proof of this theorem is due to Cauchy (1789-1857). 

Maclaurin's Theorem, published in 1742, was admittedly founded on 
Taylor's. This therorem had in fact been previously published by Sterling, 
in 1717. 

85. A Series is a number of terms which follow each other in 
obedience to some law. 

Series are either Finite or Infinite ; and infinite series are 
either Convergent or Divergent. 

86. Finite Series. A series is finite when the number of its 
terms is finite. Thus, in the expansion, 

(x + i) 4 = x 4 + 4 x z -\- 6x? -f- 4 x -f- 1 , 

the second member is a finite series. 



106 Differential Calculus 

87. Infinite Series. A series is infinite when the number of 
its terms is infinite. Thus, in the expansion, 

= i + x + x 2 4- x s + x* + x 5 + • • • x n ~ x -J----, 



I — X 

the second member is an infinite series. 

88. A Convergent Series is an infinite series, the sum of the 
first n terms of which is a variable whose limit is finite. 

All other infinite series are divergent. 

Thus, in the geometrical progression given above, we have, 

from algebra, for the sum of the first ?i terms, 

i — x n 
i + x 4- x 2 -f- x s + x 4 -f- x T ° + • • • x n ~ x = 



I — X 

I X 11 



I — X I — X 

Hence, if x < i, we have 

""i — x r 



Limit 



i + x -\- x 2 + x 3 + • • • x n ~ x 
— a finite quantity. 



= Limit 



i — x 



i — x 

Therefore it is a convergent series. 
If x > I, 

Limit i + x + x 2 + x z + • • • x n ~ 1 

= Limit = oo. 

\_i — x J M=00 

Therefore it is a divergent series. 

89. Definitions. — The sum of a finite series is the sum of its 
terms. The sum of a convergent series is the limit to which the 
sum of the first n terms approaches as n is indefinitely in- 
creased. 



Series 107 

The Remainder after ?i terms is the difference between the 
sum of the series and the sum of its first n terms. 

Obviously, in a convergent series, this remainder is a variable 
whose limit is zero as n is indefinitely increased. 

Thus, in the progression given above, 

Remainder after n terms ; 



i — x 

and if x < i the series is convergent, since 

T x n ~ 
Limit = o. 



|~ x n 

L 1 — X \n 



90. The development of a function consists in finding a series, 
the sum of whose terms is equal to the given function. Since 
any given function is necessarily finite, the sum of any equiva- 
lent series must be finite ; hence the term develop7nent applies 
only to finite and convergent series. 

91. Methods of Development : 

I. By algebraic processes. 

> 
i. By division, as 

= i + x + x 2 + x s + • • • x n + ■ ■ ■ 



I — X 

x n — y 



= x 71 -^ + x n - 2 y + x n ~y + • • • f 



x — y 

2. By involution, as 

{x + yf = x s + 3 x?y + 3 x f + f • 

3. By evolution, as 

V a 2 -f- 2 ax + x 2 = a? + f a~ 3 x — ^ a~* x 2 -}-••• 

II. By general formulae : 
1. Maclaurin's formula, ' 

fix) = f(0) + f'(0)x + f (0)j| + r"(0)| 3 + • • • etc. 



108 Differential Calculus 

2. Taylor's forniula, 
fix + y) = fix) + f'(x)y + fix) ^ + fix) | 2 + • • • etc. 

Note. — There are various other formulae more general than those just 
given. The limits of this work preclude their discussion here. It may be 
remarked in passing that as yet no perfectly general method of distinguish- 
ing convergent and divergent series has been discovered. Reference will 
again be made to the subject at the end of the chapter. 

92. Maclaurin's Theorem. — The object of Maclaurin's Theorem 
is to develop a function of a single variable i?ito a series arranged 
according to the ascending powers of the variable. 

Let f(x) = A + Bx + Cx 2 + Dx z + • • ■ (i) 

be the proposed development, in which A, B, C, £>, etc., are 
finite constants, whose values are independent of x. It is re- 
quired to find the values of these constants. Writing the suc- 
cessive derivatives of f(x), we have 

f\x) = B + 2 Cx + 3^+ • • • , 

/"(*)= 2 C+ 2. 3 . Z>T-f- • • • , 

f'"(x) = 2; 3 .Z>+ • • • , etc. 

Since the constants are independent of x, their values will be 
unaffected if we make x = o. Hence, making x = o in the 
functions, and its derivatives, we have 

A =/(o), 



C = 



2 



D= f -^1, etc. 
Substituting the values of the constants in (i), we have 

fix) = /(0) + f'(0)x + f'iO) | + f"iO) |j + • • • (2) 
for Maclaurin's formula. 



Series 



109 



93. Remark. — Maclaurin's formula fails to develop a func- 
tion of a single variable in the following cases : 

(i) When it leads to a divergent series. 
(2) When the function, or any one of its derivatives, 
becomes infinite for x = o. 



EXAMPLES. 
1. Develop (a -f x) 5 
Here f(x) = (a 4- xj* 
f(x) = 5 (a + xy 
f'(x) = A . S .(a + xy 

/'"(*) = 3-4-5 + *T 
f"(x) = 2.3.4.5(0 + *) 
/ v (*) = 2.3.4.5 

Since /(*)=/(o)+/'(o)*+/>)r +/'"(o) p + ... 

we have 

(0 + tf) 5 = # 5 + 5 <7 4 # +10 <7 3 a- 2 +10 a 2 x s + 5 0** -}- # 5 . 



/(o) = 5 , 

/'(°) = 5 ^ 
/"(o) = 4-5 «*, 

/'» = 3-4-5 A 

7 iv (°) = 2 -3-4-5^ 
/ v (o) = 2.3.4.5, 



AT 



.Y u 



2. Develop 
Here 



— (1 — #) 



— 1 



1 — x 

/(X) = (I - *)-* 

./» = (1 - ^- 2 

/"(*) = 2(1- x)~* 

f"\x) = 2. 3 .(l-x)-* 

f\x) = 2.3.4(1 -*)~ 5 
etc. 



/(o) = i, 
/'(°) = 1. 

/» = [£, 

/'» = la, 

/ lv (o)=l4- 



etc. 



Since 
we have 



ar 



/(*) =/(o) +/'(o) * +/"(o) r; +/'"(°) 



\l 



x° 

5 



+ 



.V 



= 1 + # + x 2 + * 3 + 



110 



Differential Calculus 



3. Develop sin x. 

Here f( x ) = sm x 

fix) = cos x 
f"(x) — — sin x 
f'"(x) = — cos x 
f lv (x) = sin x 
etc. 



/ (p) = sin = 0, 
f'(o) — cos = 1, 
f"(p) = — sin o = o- 
/'"(o) = _ cos O = - I, 
f lv (o) = sin = 0, 
etc. 

,3 



Since f(x) =/(o) +/'(<>) at +/»,- +/"(<>),- 

L? I_3 



y« /y& yy»7 

we have, sin x=o-\-x-\-o — -. — h o + i ho — -7— _j 

[3 [5 

»/V «/V i^V 

[3 Li Lz^ 



Lz 



It will be observed that f [y (x) = sin x =f(x); hence the 
5th, 6th, 7th, and 8th derivatives will be the same as the 1st, 
2d, 3d, and 4th respectively. We are therefore enabled to 
extend the development indefinitely without further differentia- 
tion. This development, together with those for cos x, sin -1 x, 
cos - x x were given by James Gregory in 1667. 

wV %\r V\ 

4. Prove cos x = 1 — -. — hi tt A- • • • 

[2 [4 [6 T 

5. Develop log a (1 -f- #)> » being the modulus of the 
system. 



f{x) = log a (i + *) 
f'(x) = m (1 -f -*) -1 
/"(*) = - m (1 + *) -2 
/"'(*) = 2 ** (1 + *)- 3 

/ iv (V) = -6«(i+ x)~ : 
etc. 

.-. log a (1 + x) = m (x - 



/(O) =l0g a I = 

/'(o) = *, 

/» = - «, 
/'"(o) = 2 m, 

/ iv (o) = - 6 ^, 
etc. 



x 



--- + --...)■ 

2 3 4 5 



w 



Series 111 

If a = e, m = i, 

-* i>j /yiO /v"^* yy»0 

.*. log Cl + #) == X 1 1 ~~~ •• ( 2 ) 

2 3 4 5 

(i) and (2) are called the logarithmic series. Nicholas 
Mercator published this series in his Logarithmotechnia in 1668. 
It was the first series published. 

6. Develop a x . 

Here a* = 1 + log a . x + log 2 a l-log 3 ^, — h • • (3) 

If a = e, we have 

x^ X / \ 

**=I +* + ,- +T-+ (4) 

Series (3) and (4) are called the Exponential Series. 
If in this last equation we make x = 1, we have, 

111 1 

* = 1 + 1 H r-^H 1 V . . . 

2 24 120 

.*. ^ = 2.718281, 

i.e., the base of the Napierian or Hyperbolic System of 
logarithms. 

7. Develop log x. 

Here f( x ) = &gx •'• f(p) = — 00 . 

It is unnecessary to proceed further. The function cannot 
be developed by the theorem. See § 93 (2). 

8. Show that the following functions cannot be developed by 

Maclaurin's Theorem : 

3 1 
cot x, esc x, x*, a x 

9. Develop sin -1 .*. 

f{x) = sin" 1 x .-. /(o) = o, 

f'(x) = (1 - x 2 )- 1 = 1 + -^ + !* 4 -}- ^|^+ . . .'. /'(o) = 1, 



112 Differential Calculus 



/-(^) = i + \ * + f x* - ■ • /. r\o) = i . 

etc. etc. 

1 i x s i.i. x 5 
.*. sin -1 .* = x -\ — 1 h • • • 



2 3 2 -4 5 

It will be observed that the process of arriving at the suc- 
cessive derivatives in this example is simplified by expanding 
the first derivative by the Binomial Theorem. 

10. Develop tan -1 .*. 

f{x) = tan -1 ^ .*. f{°) — °» 

f(x) =(i+x 2 )- 1 =i-x 2 + x 4 -x« + x 8 - ••• .\/'(o)=i, 

/'(i) = -2i|4.t 3 -6i 5 +8i 7 ---- .-. /"(o)=.o, 

f"(x) = - 2 + I2f- 3 X*'+ • • • .'. /'"(o) = - 2, 

/ iv (#) = 2^X — I20^ 3 + • • • .'. / iV (o) = O, 

.•. tan _1 x = x 1 f- ••• 

3 5 7 

If x = i, we have, 

i T I I I I _, 

tan -1 1 =-=i 1 1 ••• = .78^08 

4 3 5 7 9 ' 

Derive the following, 

2 ^ 3 

11. ^sec # = 1 +^-f-^rH h • • • • 

3 

_i x 2 x 3 

12. ^ tan x = 1 + # H — — 

2 6 

13. ^ ina: = 1 + x H ^-+ •••- 

2 8 



Series 113 



14. log (i + sin x) — x Y~T — ■ h 

sv } 26 12 

15. log(l+V)=log2+^ + J-^ + 

4 

16. ^ si »*= 1 -f-^ 2 -f- — -f .... 

3 



94. Euler's Exponential values of sine and cosine. 
If in the exponential series, Ex. 6, 

,y»2 yO .y.4 

r? ll [4 

we substitute for at, .# V— i and — # V— 1 successively, we 
have 

li [4 ( [3 Is 
<f V ^ 1 = I — .— + . • • • — V— I \ X — ,- + , • ■ • 



li li 

.*. Examples 3 and 4. 



^ v x = cos x + V— 1 sin #, 
<? -a: v_1 = cos a: — V— 1 sin ^. 



Hence sin x = 



2 V— 1 

V=ii „-a:V=l 



cos # = 



95. Taylor's Theorem. Taylor's Theorem has for its object the 
expansio?i of a function of the algebraic sum of two variables into a 
series arranged according to the ascending powers of one of the 
variables. 



114 Differential Calculus 

Let f(x + y) = A + By + Cf + Df + etc., 

be the proposed development in which A, B, C, D, etc., are 
functions of x and the constants which enter the function. It 
is required to find the values of A, B, C, D, etc. Since the 
proposed development must be true for all values of x and y it 
will be true for any value a of x. Let A' ', B' ', C, D' , etc., be 
the values of the coefficients for x = a ; then 

/(a + y) = A' + B'y + Cf + D'f + etc. . . (a) 

Writing the successive derivatives with respect to y, we have, 

f{a +y) = B' + 2 C> + 3 £/f + etc., 
/ ; ' (a+ } ) = 2 C + 6 B'y + etc., 

/"'("+ J) = 6 ZX + etc., 
etc. 

Since the original function as well as its derivatives must be 
true for all values of y, they are true when y = o ; hence 

/(a) = A', 

r (o = b, 

2 

etc. 
Substituting these values in (a), we have, 

/(a + y) =/(a) +/' {a)y +/" (a) | +/'" (a) | + etc., 

for the proposed development when x = a. But a is #/ry value 
of x\ hence, generally, 

fix + y) =f{x) + f'(x)y + f"(x) | + f'"(x) g + etc. . (i). 



Series 115 

Cor. i. If in (i) we make x = o and change y to x, we 
have, 

/(x) =/(o) +/'(o)x +/"(o) J +/'"(°) r + etc -> 

Li L3 

which is Maclaurin's formula. Hence Maclaurin's formula is a 
special case of Taylor's more general formula. 

96. Remark. Taylor's Formula fails to develop a function 
of the sum of two variables in the following cases : 

(i) When it leads to a divergent series. 

(2) When the function or any one of its derivatives becomes 
infinite for a value, or values, of one of the variables, it fails for 
that value or those values. 

EXAMPLES. 

1. Develop (x, + yf. 

Here f(x) = * 5 , 

f\x) = 5 x\ 
f"(x) = 4.5.^ 

/"'(*) = 3.4.5.^ 

f\x) = 2.3.4.5.*, 
/ v 0) = 2 -3-4-5- 



Since,/(* + y) =f(x) +f'(x)y +/"(*) £ +/'"(*) f- + • • 

f y 3 

we have, (x + yf = x> + 5 x 4 y + 4.5 x 3 .— -f- 3.4.5 x 2 — 

XL \2l 

+ 2.3.4.5*- + 2.3.4.5-. 

L4 [5 

= x 5 + 5 * 4 / + 10 *y + 10 *y + 5 ■#/ -1- j 5 . 



116 Differential Calculus 

2. Develop a x+y . 

f(x) = a*. f'"{a) = log 3 a.d*. 

f\x) = log a.a x . f\a) = log 4 a.a x . 

f f, (x) = \og*a.a x . 

f\d) = l g n a.a x . 

Since, f{x +y) =/(x) +f'(x)y +/"(*)]? +f"\x)£ + • 

y y 3 

we have, a x+y = a x + log #.tf x _y + log 2 ^.« x . f- + loefa.a*;— 

\l [3 

+ • • • log n tf.tf x .^- + • • • 

3. Develop sin (^ -f- 7) and prove that 

sin (x -\- y) = sin # cos _y 4- cos x sin _y. 

Here f(x) = sin ^. f' n (x) = — cos #. 

f'(x) = cos jc. f lv ( x ) — sm ■*• 

f'\x) = — sin jc. 

v 2 y 3 

.*. sin (x -|- y) = sin 3: + cos .#.y — sin x. cos x\— + • • 

\l [3 

/ y y \ iff 

= sin * 1 — f- + f • • • + cos # [y — f- + • 

V [2 Li / V [3 5 

Hence, Exs. 3 and 4, p. no, 

sin (x -\- y) = sin .# cos y -f- cos # sin j. 
Prove similarly the following trigonometric relations : 

4. sin (x — y) = sin x cos _y — cos x sin _y. 

5. cos (x -\- y) = cos ^ cos y — sin jp sin y. 

6. cos (jp — y) = cos jp cos y -f- sin jc sin y. 



Series 117 

7. Develop log a (x + y), m being the modulus of the system. 

If x = i , we have, after replacing y by x, 



log a (i + x) = m \x -- + -- — + --" ') * * (<0 

an expression previously deduced by Maclaurin's theorem. 
See Ex. 5, p. no. 

If x = o, the formula fails to develop the function. § 96, (2). 

If m = 1 in (a), we have, 

■ y. » */v ••V »^V vV 

log (1 + X) = # 1 1 

2345 

Derive the following : 

8. e* + * = Ai +y+f + r + ' 

V il [3 

9. log sec (^ -f- y) = log sec jp + tan x.y +' sec 2 # — 

+ sec 2 x tan # (- 

3 

1/2 

10. sin -1 (# -\-y) = sin * .# + y (1 — ^) i+x^i—x 2 ) § — 

3 ^ 

+ (1 +2**)(i -^)-i^-+ • • • 

If x = 1, the formula fails to develop the function. § 96, 2. 

97. Bernouilli's Series. Resuming Taylor's formula 

f{x + y) =/(x) +/'(*> +/"(*)£ +/"'(f)£ + etc -> 
and making jy = — x and transposing, we have, 

/(*) =/(o) + xf\x) - |/"(*) + g/"'(*) - g>(*) + • • • 



118 



Differential Calculus 



By aid of this formula we are enabled to expand a function of a 
single variable into a series. Thus, 



x 1 



e~^ = i 



xe~ 



__ XT 
e x — , — e '" - 



x 



\1 13 [4 

Dividing through by e~ x and transposing, we have, 
i x 2 x s x* 

e [2 [3 [4 

See Ex. 6, p. in. 

98. We have referred in a previous article ^90) to the fact 
that the term development was inapplicable to a divergent 
series, and the student was cautioned not to accept an infinite 
series obtained by any of the foregoing methods as the develop- 
ment of the function, unless the question of its convergency or 
divergency had been previously settled. It remains to show 
how series are examined for divergency and convergency. 

99. Lemma. If f(xj] a = o and f(x)\ = o a.7idf(x) is continu- 
ous between these limiting values of x, then fix) = O for some 
value of x intermediate between x = a and x = b. 




Fig. 14. 



Let APB be the locus of y = f(x), and let OA = a, and 
OB = b- then 

f(xj\ a = o, and/(#)] 6 = o. 

Since, by hypothesis, f(x) is continuous between the points 
A (a,6) and B (b,o), there must be some intermediate point 



Series 119 

(OC, CP) where the tangent to the curve is || to the ^-axis. 
But at such a point 

g =/'(*) = o. §i 9 - 

Hence the proposition. 

100. Lagrange's Theorem on the limits of Taylor's Theorem. 1 
Taylor's formula may be written, — 

f{x + y) = f(x) +f{x)y +/"(*).- + • ' • +/ M_1 (^) • 



n — i 



+ Pr~ (i) 

in which P is some function of x and y, and 

P -. — = Remainder after n terms. 

\n 

It is desired to find the value of P and thence the value 

Of P f- • 

\n 
Let y = X — x. Substituting in (i) in every term, including 
P, and transposing we have 

/(X) -/(x) -f(x){x- x) -/»^^) 2 

^ t/ . (X-xY- 1 (X-xY 

in which ? is now a function of x and X. 

Replacing x by z in every term in the preceding expression, 
excepting P, and representing the resulting expression by <j>(z), 
we have 



*0) =/(*) -/(*) -/'(*) (* - ») -/"(*) 



li 



-/■-(,) (^-'y-' ^p^pg! (3) 



1 This discovery of Lagrange placed Taylor's theorem on a satisfactory basis for the 
first time . 



120 Differential Calculus 

Now if z = x, we have <f>(z)] x = o, since the second member 
of (3) reduces to the first member of (2). 

Again, if z = X, we have <^>(z)\x = °> since each term in the 
second member becomes zero. 

Hence <f>' (z) = o, for some value of z between the limiting 
values x and X of z (§ 99). Let be a proper fraction. Then 
z = x + (X — x) represents any value of z between the values 
z = x and z = X. 

Differentiating (3) we have, after cancellation, 



,ft,-y^g?£? + /.g=3 



»— 1 

- I 



Now, if 2 = a: + (X — x), we have cj>' (V)] ^ + e ( X _ ^ = o, 

hence, P=f n (x + 6(X- x)). 

y n 
Multiplying both members by -. — and changing X — x to y } 

we have *— 

j>r. =/Hx + df) r (4) 

for the remainder after n terms. Substituting this value in (1), 
we have 

/(* + y) =/(•*) +f'(*)y +/"(*) {?+••■ +/*"' (*) " 



li 



« — I 



+/»(* + <Wj£ (5) 

for the completed form of Taylor's theorem. 

y 1 

Had we assumed Py, instead of Py- > to represent the remain- 

\n 

der after n terms at the outset, and followed the course of rea- 
soning just concluded, we would have obtained 

(j —BY- 1 

jpy=/*(x + e y y > t f .... (6) 



Series 121 



as a second form of the remainder, and, consequently, 
fix +y) =/(*) +/'{x)y +/"(*)£+ ■ ■ ■ +/"-\-r) ' 



\1 



n— \ 



d — Oy-i 
+f n {x + 0y) ^ n J T r (7) 



as a second completed form of Taylor's theorem. 

If we make x = o in (5) and (7), and then change y to x, we 
have 



/(*) =/(°) +/'(°> +/"(°)n- + • • '/"-^pr 



+ /"(**) |. (8) 

and 

f{x) =/(o) -f-/'(o> +/»| + • • '/"" 1 (°)^^ 

+ /"(^)n — — * n (9) 



as the two completed forms of Maclaurin's theorem. Similarly 
from 4 and 6, we have 

/*(«*)£ ( io ) 



and 



f»(0x)" — ^ x n (11) 



n — 1 



for the two forms of the remainder after n terms in Maclaurin's 
theorem. 

Since (§ 89) the remainder after any n terms, in any conver- 
gent series, is a variable whose limit is zero, the question of 
divergency or convergency can generally be ascertained by 
examining this remainder. 

101. If the n th derivative is finite for all values of n in Tay- 
lor's and Maclaurin's expansions the fufiction is developed. 



122 Differential Calculus 

For the (n— i) th and u th derivative in Taylor's expansion we 
have 

v «-l y n 

x ' \?i — i v J \n 

Dividing the latter by the former, we have 

/»(*) y 

Now /" (x) is, by hypothesis, finite, and, since the value of n 
is arbitrary, J n ~\x) is also finite ; hence 

r(x) 



/»-!(*) 



is finite. 



y 

Again, since y is necessarily finite, - becomes ultimately very 



small, and approaches zero as a limit as n approaches oo . 

If- 1 ' 



\ x ) y 

Limit 



(x) 



= o. 

n = oo 



Hence the values of the terms become infinitely small as n 
becomes infinitely great, and their sum, after n terms, ap- 
proaches zero as a limit. Hence the series is convergent and 
the function developed. 

The proof is similar for Maclaurin's expansions. 

102. Let us now examine a few of the expansions previously 
determined, and ascertain if the term development used in con- 
nection with them has been properly used. 

1. sin x = x — -j— hi !— + •"'■ Ex. -z, p. no. 

li li 17 

Examining the derivatives, we see that 

f n (x) = ± sin x, or ± cos x, 

according as n is even or odd. Hence the n th derivative is 
finite for all values of n, .'. (§ ioi) the function is developed. 



Series 



123 



Similarly we may prove that 



/y^i *yr& •y^D 

cos x = i — .- — hi nr + etc. Ex. 4, p. no. 

[2 [4 [6 



>y^ 'I -y<^ 

2. log (1 + .*) = a: 1 h • • • 

234 



Ex. 5 p. no. 



In this case, f n (x) = (— 1) 



»— 1 



\n— 1 
(1 + ^) w 



Hence, / n (0*) r- = ^ 1 , 

J v J \n n V 1 -V ® x 

Hence, 



Limit 



f n (Ox) 



x 71 ' 
\n_ 



= Limit 

n= oo 



"(-0 



M — 1 



« 



# 



i -f &%• 



= o, 

72= 00 



if x < + 1; •'• the function is developed for all positive values 
of x less than unity. By using the second form of the re- 
mainder we can show that the function is developed when x 
lies between o and — 1. 



x 2 
3. a x = 1 -j- log a . x + log 2 a \- 



li 



Ex. 6, p. in. 



Here, 



f n (x) = a? log" a. 



Hence, / B (Ox) r- = a^ log" a t~ = ± — ^-^ c 

J v J \n s \n \n 



9x 



Since a 0x is finite and limit 
we have, 



~(x log d) n 



Limit 



~(x log a) n 



IT" 1=. = °' § io1 ' 



Hence the function is developed. 



^ "Z Xr 

4. Sin -1 * =1 + - + — h • ♦ • Ex. o, p. in. 

6 24 5 

Here the ratio of the n th term to the (n — i) th term is 



124 Differential Calculus 

where c is some function of // and a finite quantity when n 

xF\ 
is finite. Necessarily x < i, .*. Limit — = o; 

.*. the series is convergent. 

5. tan -1 .* = x 1 . 

3 5 7 

The reasoning is the same as in Ex. 4 when x < 1. When 
x > 1 the series is divergent. 



Illusory Forms 125 



CHAPTER IX. 

ILLUSORY FORMS. 

History. — The Marquis de St. Mesme (L'Hospital) published in his 
calculus (1696) a partial investigation of the limiting value of the ratio of 

functions which for a certain value of the variable take the form — • 

o 

John Bernouilli, the elder (1667-1748), was the first to solve the problem 

by aid of the calculus. 

103. We are accustomed to consider the value of a fraction 
as indeterminate when for any given value or values of the vari- 
ables which enter it, it assumes the form - , — , o . 00 , etc. 

o 00 

It is our purpose to show (1), that such expressions are not 
necessarily indeterminate but are frequently illusory, and (2), to 
indicate a method by means of which their true values may in 
general be ascertained. 

104. Definition. The value of a fraction is the limit it ap- 
proaches as its numerator, or denominator, or both, approach an 
assigned value or values. 

x 2 1 

Thus, the fact that ■ = - when x = 1 may be ex- 

3^—12 

pressed more generally, x being a variable, by writing 



Limit 



x 2 



_3 x ~ J _ 



a;=l 



105.* Evaluation of the forms 

a o 

* Kepler introduced name and notion of i7ifinity into geometry in 1615. 



126 Differential Calculus 



I. Form - 



o 



Here, Limit 



x 



a = 



a 

- = 00 
o 



i.e., when the denominator of a fraction becomes infinitely 
small and approaches zero as its limit, the numerator being 
constant and finite, the value of the fraction becomes infinitely 



large. 






II. Form - . 
a. 




Here, 


Limit 


a 



x= 



o 
a 



i.e., when the numerator of a fraction becomes infinitely small, 
and approaches zero as its limit, the denominator being con- 
stant and finite, the value of the fraction becomes infinitely 
snail, and also approaches zero as its limit. 

III. Form - . 
o 

This case frequently admits of evaluation, 

{a) By Algebraic or Trigonometric reduction. 

,% — 7 X -\~ I O O 

Thus ~ = - when x = 2 ; but on factoring and 

XT — 4 O 

cancelling, we have 

X 2 — 7 X -f- IO (x — 2) (x — X — K 2 , 

— — - when x = 2. 



x 



4 (x — 2) (x + 2) X + 2 



X o 

Again : — — = == = - when x = o ; but multiply- 

\la + x — Na — x ° 
ing both numerator and denominator by the complementary 

surd \/a + x -J- \ja — jp and reducing we have, 



Illusory Forms 



127 



X 



Vtf + x + \Ja — x 



\Ja + x — Va — 



= ya when x = o. 



# 



. . COS 2 .%• O , 7T , 

Again : = - when x = - : but 

cot 2 x o 4 

cos 2 a: tt 

= sin 2 # = i when ^ = - . 

cot 2 # 4 

Obviously these methods are special in their character, and 
apply only to those algebraic and trigonometric forms which 
admit of ready reduction. We have, however, a general process 
afforded, 

(J?) By the Differential Calculus. 



To deduce this process let 



<f>(x)' 



<K«) 



\\> x \ a ^ (a) o 
that is, let <f> (a) = iff (a) = o. 

By Taylor's theorem, we have, 

*(*)+*'(*)j> + *"(*)£ + ■ 



* ( * + f) iff (x) + $ (x)y + iff" (x)(- + • • • 

Li 

Making jc = #, we have (since (f> (a) = iff (a) = o and y is a 
factor of both numerator and denominator), 

*'(*)+*'>)■!++"'(*)£+ ... 

4>(a+y) _ \1 13_ 

*P(a+y) " y y* 

iff (a) + iff (a) .- + iff (a) .- + • • • 



Hence, 

<f>(a+y)' 



Limit 



iff(a-\-y)_ 



= Limit 



2/=0 



12 [3 

Li 13 



w 



?/=o 



128 Differential Calculus 

<f> 0) V (a) 



i.e. 



Hence ^ 



f(a)" f (a) 
*' (x) 



fW 



1- 



d> (x) o 
Hence the general rule, if = - for a?iy value a of x the 

if/ (x) o 

value of the fraction may in general be found by dividing the first 
derivative of the numerator by the first derivative of the denomi- 
nator and then substituting a for x in this ratio. 

<j>(x) .x 2 — 7 * -f i o o 

Thus ; ; = zf = - when x = 2, 

if/ (x) xr — 4 o 

<f> f (x) 2 x — 7 3 

.-. T7-7-T = L = — - when ^ = 2. 

j// (x) 2 x 4 

Hence, as before III (a), we have — - for the true value of 
the fraction. 

Cor. 1. If , ; ^ = - i.e., if <*/ (a) = \1/ (a) = o, then the 

xp (a) o v 7 

i- • j • J r ■ / x 0) *"(*) 

limits determined from equation (1), are — ^4 = „ ' • 

If cf>" (a) = \\i" (a) = o, we have similarly 

<f>(a) <t>"'(a) 



Hence, generally, 



xp(x)_ 



tf{x\ 



^Ml = etc 
when these forms successively assume the illusory form - 






EXAMPLES. 

^ -4- ^ x — 2 
1. Evaluate the fraction when x = o. 

I — COS X 



Illusory Forms 



129 



Here 



xp(x)_ 

i// (x)j 

4>" fa) 
i//' fa)_ 



e° -\- e x — i 
i — cosx 

e* — r 



sm x 



o COS X 

e* -\- e~ x — 2 



I — COS X 



= 2. 



1" 



Evaluate the following : 



2. 
3. 



JC — 


2 




(x- l) 
x 3 — I~ 


»_ x 


X — I 


1 


# — sin 



X 




/sin /zjc 


\m ' 






X 



7. 

8. 

9. 
10. 
11. 
12. 



i — sin x + cos # 
sin x + cos ^c — i 



a™ x — 


a 




log sin ^c 


7T * 
2 


^ — 


e~ x 




log(i 


+ x)_ 





xf° 


— x 




I — X 


+ log X 


1 


e* — 2 


sin x — e~ x ~ 




X 


— sin x 





X 4 — 


2 X S -\- 2 X — I 



.V" 



15 r+ 24a: - 10 



Ans. - . 
11 

Ans. 3. 

Ans. log- . 

Ans. — . 
6 

Ans. 1 . 

^4;?^. a log #. 

^;w. 2 . 

Ans. — 2. 

y4/2.5\ 4. 
^4/2.T. . I 



130 



Differential Calculus 



106. Evaluation of the forms — » — » — 



oo 



00 



I. Form 



oo 



Here, evidently, Limit 



= o. 



II. Form 



oo 



Limit 



= oo 



III. Form 



oo 
oo 



Let 



<M>) 
xp(x) 

i.e., let <f> (a) = \j/ (a) = oo . 



1 *00 



CO 
CO 



Taking the reciprocal of the numerator and denominator 
we may write, 

i 
<J>(x)~ 



'/'OO. 



xp(x) 



<f>(x) 



By § 105, III. (fr), we have, 



1 

if/ x 
1 

<f>(x) 



hence, 



iff (a) 



[*(*)]' 

- 4>'Q0 

= ~ <frO0 " 

" ^ oo. 



a 


jAOO. 


2 fW" 
0' 00_ 



f(d) m 






(0 

0) 



i.e., 



<£00" 



<£'O0~ 



«AO0Ja «A'00_ 



Illusory Forms 



131 



Hence, the rule for evaluating the fraction 



<K*) 

xfj(x) 



that takes 



CO 



the form — for any value a of x is the same as when it takes 



00 



the form -. See § 105, III., b. 



Thus, 






1- 



log sin x 



cot x 
cos x n 



00 
00 



<£'(V)~| sin x 

$' {x)\ - csc 2 ^j 

log sin x 



= — sin x cos x~\ = o. 



cot X 



= o. 



tion (1) of the same article we divided through by 



107. Remark. — In deriving equation (2) Art. 106 from equa- 

jp(a)_ 

viously if the real value of this fraction is either o or 00 the 
generality of equation (2) is not established. Let us examine 
these cases. 

I. When *^> = o. 

Let c be any constant, then 

<f> (a) <j> (a) -J- c\p (a) 00 

if/ (a) if/ (a) 00 

. since <j> (a) = if/ (a) = 00 . But the real value of this expression 
is c\ hence 

£0) , c = . <£ (a) + cxp (a) = # (a) + cip'(a) = f(a) + ^ 
if/ (a) if/ (a) if/' (a) fA'M 

. <f>(a) 4>'(a) q 

Hence Equation (2) is true in this case. 



132 



Differential Calculus 



II. When 



Then 



Hence, by I., 



<M>) _ 

<K g ) ~ <i> (a) 



00 . 



= o. 



= o 



00 . 



and equation (2) holds in this case. Hence, equation (2) is 



generally true for the illusory form 



00 



00 



We may write, therefore, generally, 



<f>(x)~ 



*'(*)" 



«/y(V>J« xp r (x)\ a f (*)J 



_*"(*)■ 



,„) ; = etc. 






00 



when the fractions assume successively the illusory form -^. 



EXAMPLES. 



1. Evaluate 



log x 





I 


X 


<J>(x)~ 


log X 


xp(x)_ 


I 


X 


I 


*'(*)" 


^ 



when x = o. 



00 
00 



-Jo 



1// (*)_| 



= — x\ = o. 



Illusory Forms 



133 



Evaluate the following : 



2. 



3. 



4. 



log 

cot 

I 

X 



cotx 
sec 



xl 



Ans. o. 



Ans. i . 



sec 3 



x 1 



Ans. — 3. 



It will be found simpler to transform this fraction so as to 
make it take the form §. 



log* ! 
x 

So 



Ans. o. 



108. Evaluate the forms o • 00 and 00 — 00 . 

Expressions which assume these forms can be readily reduced 
to expressions that assume either of the forms % or °9_. 



EXAMPLES. 



1. Evaluate x log x when x = o. 
Here [x log x] = o (— 00 ). 



But \x log x] = 



~\ogx 



00 
00 



Hence, Ex. 1, p. 132, we have 

x log x = o when x = o. 
2. [sec x — tan x] n . 

Here [sec x — tan #] „. = 00 — 00 . 

2 

_ r ., 1 — sin 

But [sec x — tan x\ n = 



-1 



II " °" 



134 



Differential Calculus 






— COS X 

— sin x 



= cot xl = o. 

7T J 77 



3. 
4. 

5. 
6. 

7. 

8. 
9. 



[sec x — tan oc\ n = o. 

2 

[2 # tan x — 7r sec .#]„.. 

2 

[sec 2^(i- tan .#)]„.. 

Li L_l. 

log.* X — ij! 

I Jt: ~| 

log* logjcji 

. N 7TX~\ 

(1 — x) tan — • 
2 Ji 

|>— log A^. 

[sec 3 a: cos 7 x] n . 

2" 



^;z,r. — 2. 
y4;/j-. 1 . 



Aiis. — 1, 

7T 
-<4^. O. 



^4/W. \. 



109. Evaluate the forms o°, 00 °, i 00 , o", 00 ". 

Let _y = u v , in which z^ and v are functions of .*. 
Applying logarithms, we have 

log y = v log &. 

For some value a of x, let us suppose 

(1) u v = o°, i.e., z/ = v = o ; then log y = — o • 00. 

(2) z/ v = oo°, i.e., u = 00, ?/ = o ; then log _y = o «oo . 

(3) z/ v = i°°, i.e., « = 1, v = 00 ; then log y = 00 • o. 

(4) ?/ v = o 00 , i.e., ?/ = o, z/ = 00 ; then log y = — 00 • 00 = 

— 00, .*. y = o. 

(5) u v = oo°°, i.e., u = 00, v = 00 ; then log _y = 00 • 00 = 00, 

.-. y = 00. 



Illusory Forms 



135 



It appears, therefore, that forms (4) and (5) are not, properly- 
speaking, illusory, and that the logarithms of forms (1), (2), and 

(3) may be evaluated by the method explained in the last article. 
1 1 

Thus (1 + ocf = 1 °° when x = o. Let y = (1 -f- x) x , and, ap- 
plying logarithms, we have 



log y = — log (1 + x) = 00 • o when x =^ o. 



But 



x 



log (1 + X 



)].-'^ 



+'*)" 



_ 



Hence, 



= 1. 



f(x)~\ = 1± 
^'OOjo 1 

,\ log 7 = I. .'. y = e. 

i.e., (1 + *) x Jo = ^. 
Again, ^Jq = o°. Let 7 = x r , then 

log y] x=0 = x\og x] == — o • 00 = o. See Ex. 1, p. 133. 
.-.Iog7=o. .-. y = A^jp = 1. 

110. Evaluation of compound illusory forms §•§, §•£, etc. 

Such forms arise from factors in the function that take illu- 
sory forms for the given value of the variable. 

We may evaluate the function by evaluating the factors sepa- 
rately. Thus 



sin 2 ^(sec>r — i)~| /sinjcV 1 — cos x 



x° 



l-( 



x 



X COS X 



o o 
° o 



= 1; 



_. ; sin.r~] cos^l /sin.x\ 2 

But ~r = ^ =I - •'• hr) 

**- Jo x Jo \ * / J( 

, 1 — cos x~\ sin x 

d = ,— \=% = o. 

x cos x J G cosjc — ^sin^J 

sin 2 x (sec x — i)l /sin jcN 2 / 1 ~~ cos #Y| 
** \ x 1 \ •* cos x /Jo 



136 



Differential Calculus 



EXAMPLES. 



Evaluate the following : 

1. x*\ . 

2. x sinx ] . 



3. (i + 

4. (i + 



5. (i + ax) x \ . 

tan x — x~\ 

6. : • 

x — sin xj 



7. 



8. 



9. 



Vjp tan x 



tan #1 

X — 2 X~\ 

(f-?f— J; 



^ — ^ 



# — sin 



in * xl 
v z x J 



sin° 

10. # m (sin x) iAnx ] 

11. (sin *) tanx ] . 
^ log (i + #)" 



12. 



log (i + x) l _ 

I — COS X J 



13. x<*] . 
14. 



15. 



16. 



i — x + log # 



}■ 



■2*1 

tan # — jp J 
tan (# + #) — tan (# — #) "I 



Ans. i . 

-*4/w. i . 

Ans. e. 

Ans. i. 

Ans. e a . 

Ans. 2. 

^4/2«5\ i . 



Ans. J. 



-4«j. — £, 



^/w. i . 
Ans. 2. 

Ans. — 2 

Ans. i . 

i + <z 2 



tan 1 (a -\- x) — tan 



Ans. 



cos' a 



Maxima and Minima 137 



CHAPTER X. 

MAXIMA AND MINIMA. 

History. — Kepler (i 571-1630) was the first to observe that the incre- 
ment of a variable was evanescent for values infinitely near a maximum or 
minimum value of the variable. This remark contains the germ of the rule 
given by Fermat (1601-1665). 

The correct theory of Maxima and Minima was first given by Maclaurin 
in his Treatise of Fluxions (1742). 

111. Definitions. — A maximum value of a function is a value 
greater than the values which immediately precede and follow it. 

A minimum value of a function is a value smaller than the 
values which immediately precede and follow it. 

From these definitions it appears that a maximum value of a 
function does not mean the greatest value of the function, nor 
does a minimum value mean the smallest value of a function. 
In fact, a given function may have several maximum values and 
several minimum values. 

112. Conditions for Maxima and Minima. 

L,etf(x) be a function of an increasing variable x. Then, by 
definition, f(x) is an increasing function just before it reaches a 
maximum value, and a decreasitig function immediately after it 
passes through that value. Hence, (Art. 13, Cor.), f'(x) is 
positive (+) before and negative ( — ) after f(x) attains a maxi- 
mum value ; hence, 

f(x) = o, orf'(x) = 00 

at a maximum point, since f r (x) is continuous, and cannot, 
therefore, change sign from + to — without passing through 
one or the other of these values, 



138 



Differential Calculus 



Again, by definition, /(V) is a decreasing function of x just 
before it reaches a minimum value, and an increasing function of 
x immediately after it passes that value ; hence, f\x) is negative 
(—) before and positive (+) after/"(V) attains a minimum value. 
Hence, at a minimum value, 

f'(x) = o, or f'{x) = oo. . 

It appears, therefore, that the essential condition for a maximum 
or minimum value of a function of a single variable is that its 
first derivative shall change sign, — in case of a maximum value, 
from 4- to — ; of a minimum value, from — to +. It also ap- 
pears that the value, or values, of x which render /"(V) a maxi- 
mum or minimum will be found among the roots of the equa- 
tions formed by equating the first derivative to zero or to infin- 
ity. These roots are called critical values of the variable, and 
must be separately examined, in order to ascertain which, if 
any, give rise to a maximum or minimum state of the function. 

113. Illustration. — Since f r (x), when considered geometri- 
cally, always represents the slope of the tangent to the curve 
y = f( x ) (§ J 9)> tne principles of the preceding article may be 
graphically represented. 

I. Critical values which render f'(x) = o. 

Let SM (Fig. 15) be the locus of the equation y=f(x). 




Fig. 15. 

At the highest and lowest points of the curve, D and E, and at 
such points as F where the direction of curvature changes, the 
tangents are || to the Jf-axis ; 



Maxima and Minima 139 

hence, dy ... x 

for the critical values OA, OB, and OC, of jc. 

For a value of x a little less than OA (say 6X4') the tan- 
gent at the extremity of the corresponding ordinate A'D' makes 
an acute angle with the AT-axis ; hence, fix) = -f- quantity. 
For x = OA,f'{x) = o. For x = OA" , y =A"F>", and the tan- 
gent makes an obtuse angle with the X-axis ; hence f'{x) = 
— quantity. 

Hence at a maximum point, as D, fix) passes through o 
from -f- to — direction. Similarly, to the left of the mini?num 
point E the tangent makes an obtuse angle with X, while, to the 
right of it, the tangent makes an acute angle ; hence f (x) at a 
minimum point passes through o from — to + direction. 

At such a point as F, while f\x) = o, yet it does not change 
sign as x passes through the critical value OC,f'(x) being posi- 
tive + (or negative — ) on both sides of F\ hence CFis neither 
a maximum nor a minimum value of y. This fact is also evident 
from the definitions, since CF is neither greater nor smaller than 
the ordinates which immediately precede and follow it. 

The illustration further emphasizes the fact, that all the roots 
of the equation f\x) = o do not necessarily correspond to 
maximum or minimum values of /(x). 

II. Critical values which render f\x) = oo. 
Let SM (Fig. 1 6) be the locus of the equation y = f(x). 
At such points as D, F, F, where the tangents are perpen- 
dicular to the X-axis, we have 

g =/'(*) = - 

OA, OF, OC are therefore critical values of x. 

At D', f\xj\oA>= + quantity ; at D,f{x)\oA = oo ; at D" , 
f'(x)\ 0A ff = — quantity ; hence at a maximum point f\x) 
passes through oo from -f- to — . 



140 



Differential Calculus 



Similarly, at E,f\x) passes through oo from — to -f-. 

At F* f'( x )]oc = °° ; but for values of x a little less and a 
little greater than OC we find f\x) = a positive quantity ; 
hence f\x) does not change sign as x passes through the crit- 




ical value OC\ hence CF does not represent either a maximum 
or a minimum value of fix). This is also evident from the 
definition. It also appears that the roots of the equation 
fix) = oo do not necessarily render fix) either a maximum or 
a minimum. 



114. Methods of Investigation for Maximum and Minimum 
Values. 

I. By examining the given function. 

Lety*" (x) be the given function, and let a be any one of the 
critical values found by equating ^/"'(V) to o or oo , or both. 

Then, by the definition, § in, we have, h being a very small 
quantity, 

/(a) >/(« - X) 1 
f(a)>/(a + X)\ 

/(a) </(a - X) 
/(«) </(a + A) 



For a maximum 



For a minimum 



(?) 



Thus, let 
then, 



X" 



fix) = 3X 2 + 8x; 

f r (x) = x 2 -6x-\-8. 



= 6§ = a maxi- 



Maxima and Minima 141 

As no finite value of x will render f'(x) = oo , we equate it 
to zero ; hence 

x 2 — 6 x -\- 8 = (x — 2) (x — 4) = o, 
.'. x = 2 and x = 4, 

are the critical values of x. Substituting values a little less and 

a little greater than x = 2 in the given function, — — 3 x 2 -f 8 x, 

we have /(»]i = Sh 

/(x)y= 6§, 

/(*)i=6. 

Hence, 

mum value. 

Substituting now values a little less and a little greater than 
x = 4, we have, 

/0)] 3 = 6, 

/(*)]4=SJl 

/(*)],= 6|. 
Hence, 

/(*)! < /Wis 

/(*)]* < /(*)]■ 

a minimum value. 

Again let us consider the function, 

cf> (x) = m — n (x — 2) 3 . 

2 it 
Here <£' (x) = 7 • 

3 (x — 2)* 

As no finite value of ^ can render this derivative = o, we 

equate to 00 . 

2 /z 
Hence j = 00 , 

3 (X - 2f 
.'. X = 2 

is the critical value. Hence 



I .'./(*)]. = (f~3^ + 8*)1 = si 



142 Differential Calculus 

<£ (xyji = m — n, • 

4> (x)\ = m, 

<j) (V)] 3 = m — n ; 

" '"' -5 SSI > 5SS 1 ■■■ ♦<*»--<- — (— Ai-— 

a maximum value. 

II. By examining the first derivative. 

Let f(x) be the given function, and let a be a critical 
value of jc. 

Let h be any very small quantity. Hence, §112, 

f (a — H) = a positive quantity \ 
For a maximum <j /'(^) = o or 00 > (3) 

f (a + ^) = a negative quantity ) 

/"' (0 — //) = a negative quantity ) 
For a minimum <j /'(^) = o or 00 > (4) 

/"' (« -}-//) = a positive quantity ) 

To illustrate let us resume the example 

f(x) = 3 x 2 4- Sx. 

Hence, f (x) = x 2 — 6x-\-S = o, .*. # = 2, a? = 4, 

/'(*)] 2 = o, 
/'(*)] 3 = -i. 

Hence, (3), 3 x 2 -\- 8 x\ = 6§ = a maximum as be- 
fore. See I. 

Taking the value x = 4, we have, 

/'(*)] 3 = -i, 
/*(*)]« = <>. 



Maxima and Minima 143 

X 3 1 

Hence, (4), 3X 2 -\-8x = 51 = a minimum value. See I. 

3 J4 

Taking the example 

<!> (x) = m — n(x — 2) 3 , 

we have, <f> {x) = j = 00 , .'. x = 2, 

3 (* — 2 T 

2 

o 

4> (x)\ = co , 

2 

<f>'(x)] 3 = 0; 

•'• (3) ( w — n{x — 2) 5 )] 2 = ?/z = a maximum value. See I. 

III. By examining the second derivative. 

When_/(V) is a maximum, f (x) changes sign from -f- to — , 
§ 112 ; hence, f (x) is a decreasing function of x ; hence, f"(x\ 
must be negative for the critical value of a? that renders f(pc) a 
maximum. 

When /"(x) is a minimum, y' (jc) changes sign from — to +, 
§ 112 ; hence, /' (x) is an increasing function of x ; hence, fix) 
must be positive for the critical value of x that renders f (x) a 
minimum. 

Hence, if a be a critical value of .#, then 

For a maximum f" (a) = a negative quantity.. 

For a minimum f" (a) = a positive quantity. 

Let us resume the example, 

x 3 
f (x) = $x 2 + 8x. 

Hence, f'( x ) = x 9 — 6 #~-f- 8 = o, ,\ x = 2,x .= 4, 

and f"( x ) = 2 x — 6, 



: /"(*)],= - 2, .-. (J - 3^ + 8 ») 
/"(*)]«= + 2, .-. /| 3 _ 3 ^ -I- 8 *) 



2 = 6| = a max. value. 



= 5I. = a min. value. 



144 Differential Calculus 

Taking the example 

<£ (x) = m — n (x — 2)3, 

2 n 
we have, ^'C*") = 1 — °° > •'• # = 2 ; 

3 (^ - 2) 3 

2 /z 
also, 



^ V J 9 (X — 2) 3 



Hence, <£"(#)] 2 = °o ; hence, this method does not apply- 
when the critical value renders the first derivative infinite. 

For critical values that render f(x) = o, Method III. is 
usually the simplest. Frequently, however, the form of the 
given function is such as to render its second derivative difficult 
or tedious to obtain. In such cases Method II. should be 
employed. 

It frequently happens, when Method III. is employed, that 
a critical value of x reduces f" (x) to zero as well as f (x). 
How to proceed in such a case will be explained in the next 
article. 

115. Maxima and Minima by Taylor's Theorem. 
Resuming equation (1), § 114, and transposing, we have, 
f{a -Ji) -fid) < o 



f(a + X)-f(a) <oJ {b) 

as essential conditions for a maximum value ofjT(x) for the 
critical value a of x. 
By Taylor's Theorem : 

/(a - h) -/(«) = -/' (a) h +/" (a) I -/'"(«) I + . . 

f(a + h) -f(a) =/' (a) h +/" (a) ~ +/»' (a) ~ + . . (d) 

l_ \A 

If, therefore, f(d) is a maximum value of f (x) the second 
members of these equations, (V) and (d~), must be less than zero, 
i.e., negative. If h is taken sufficiently small — and we can take 



Maxima and Minima 145 

it as small as we please — the first term in the second member, 
f (a) /i, can be made to exceed numerically the sum of all the 
other terms. But these terms have different signs ; hence, 

/'(.«) = ° 

is an essential condition to be fulfilled in order that the second 
members may be negative. It is therefore an essential condi- 
tion for a maximum. 

Making f (a) = o in (V) and (*/), we have, 

/{*-*) -/w =+/'» | -/"»!+• • • . « 

f(a + X) -/(«) = +/" (a) | +/'" («) | + • • • • (/) 

Giving h such a value as to make the first terms numeri- 
cally larger than the sum of all that follow them, we have, as a 
second essential condition for a maximum (since /i 2 is positive), 

If f r '(a) is zero for the critical value a of x, then a similar 
course of reasoning shows that 

f"\a) = o, and/ iv («) < o 

are conditions for a maximum value. If J~ lv (a) = o, then 

f\a) = o, and/ vi (V) < o 

are conditions for a maximum value. 

By assuming equation (2) (§ 114), we can show, similarly, that 

f\d) = o, and/"(tf) > o 

are essential conditions for a minimum value of fix), and if 

f"{d) = o, that 

f'"{a) = o, and/ iv (» > o 

are essential conditions ; and so on. 
In general, therefore, if 



146 Differential Calculus 

f'(a) = o, f"(a) = o, f"\a) = o • • • f"-\a) = o, 
then, f(&) is a maximum if n is even andy" w (#) < o ; 
and fif) is a minimum if n is even and _/""(<?) > o ; 
and f( a ) is neither a maximum nor a minimum if n is odd 
andy w (<?) > o or < o. 

116. Practical Suggestions. — In the examples and problems 
which follow this article the following suggestions will be found 
of great service in simplifying the operations : 

(i). The critical value that renders f(x) a maximum or 
minimum will render C + Z)f(x) a maximum or minimum. 

(2). If fix) is positive then [f(x)~] n is a maximum or mini- 
mum for a critical value that renders f(x) a maximum or a 
minimum. 

lifix) is negative and n is an odd integer, then L/X-x')]" is 
a maximum or a minimum for a critical value that renders fix) 
a maximum or a minimum. If ;/ is an even integer, then [_/*(^)] w 
is a maximum or minimum for a critical value that renders f(x) 
a minimum or a maximum. 

(3). The critical value that renders f(x) a maximum or a 
minimum will render a minimum or a maximum. 

(4). The critical value that renders f(x) a maximum or a 
minimum will render \og a f(x) a maximum or a minimum. 

EXAMPLES. 

1. Examine mx? — 2 nx -f c for maximum and minimum 
values. 

Here f( x ) — mx ^ ~ 2 nx + ^ 

fix) = 2 mx — 2 11 = o. 

.•. ^ — _ is the critical value of x. 
m 

Also f"( x ) — 2 »*' 



Maxima and Minima 147 

Since f"(x) is positive, we have 

mx 2 — 2 nx -{- c\n = c = a minimum value. 

m m 

2. Examine 16 ax 3 — 6o ax 2 -f- 48 ax — \a for maximum and 
minimum values. 

As 4 <z is a constant factor we may omit it (§ 116, 1), and write 

fix) = 4.x 3 — 15 jc 2 -f- 1 2 # — 1 . 
Hence, fix) = 12 x* — $0 x -\- 12 = 12 (jc 2 — far -f- 1) 

= 1-2 (x — i)(# — 2) = O. 

.*. x =-\ and ^ = 2 are the critical values of x. 
f"{x) = 24^ — 30; 

/"(*)]i = ( 2 4* - 3°)]i=- 2 4. 

•'• f(p°) 1S a niaximum when x = ^ ; also 
/"(*)!= (24* - 3o)] 2 =i8. 

•'• yC-^) * s a minimum when x = 2. 

3. Examine (# — 4) 5 (# + 2) 3 for maximum and minimum 
values. 

Here f (x) = (x — 4) 5 (jc + 2 ) 4 , 

/'(*) = (jp _ 4 )5 4 ^ + 2 )3 + (* + 2 )4 5 (* _ 4 )4 

= (# — 4) 4 (a + 2) 3 5 4 (x — 4) + 5 (x + 2) J 
= 9 (x — I) (jc — 4) 4 (^ -f- 2) 3 = o, 
.«. x = |, .%• = 4, x = — 2, are critical values. 

As the work of obtaining the second derivative is tedious, 
let us use Method II. (§ 114), i.e., let us see how the first de- 
rivative passes through zero, as x passes through its critical 
values. 

As x passes through the value f , f'(x) changes sign from — 

to + ; 
As x passes through the value 4, f\x) remains positive, i.e., 
does not change sign ; 



148 Differential Calculus 

As x passes through the value — 2,f'(x) changes sign from 

+ to — . 
Hence, 

when x — %,f(x) * s a mm i mum ; 

when x = 4,J~(x) is neither a maximum nor a minimum ; 

when x =— 2,J~(x)is a. maximum. 

4. 

4. Examine b + (x — a) 3 for maximum and minimum values. 
Here f\ x ) = 3 (x — tf) 3 = o; .*. # = a. 

As jc passes through the critical value a, f'{x) passes through 
zero from — to +. .*. for x = a,f(x)\ a = &> a minimum value. 

5. /(#) = Z> + (x - af. 

Here f\ x ) = % (x — ay = o ; .\ .# = a. 

As 3: passes through the critical value a, J~(x) does not 
change sign ; hence, J~(x)~\ a = l>, neither a maximum nor a 
minimum. 

6. f(x) = 2 x s — 9 ax 2 +12 a 2 x — 4 <z 3 . 

y*(V)] a = a 3 , a maximum ; y*(V)] 2a =: o, a minimum. 

7. /(*)=- + 



+ £) 2 . . -, .-, - £) 2 

f(x)\ a* — -' a minimum ;f(x) J fl2 = ^-, 

a+b a — b 

a maximum. 

8. f(x) = ^ (;/z + xf (m — xf. 

f{x)\m, minimum ; f{x)\ _ m , minimum ',f(x)~\-m, a maximum. 

1 

9. f(x) = x x . 1 

_/"(#)]«,= <? e , a maximum. 

10. f(x) = • y ( x )\e = ~' a maximum. 

X c 



Maxima and Minima 149 

11. f(x) = cos 3 .* sin x. 

/(^)]sin-H = /(^)]^= ± A ^3> a maX « aIld a min ' 
6 

12. J~(x) = sin 3 .# cos .*. 

y"(^)]cos-H == /(^)k == ± A ^3> a max - and a min- 
is. f{x) = 



i + x tan .# 

COS X 

/(*)]cos* = — ; — = — ' a maximum. 
u i + sin.* 

14. f{x) = sin a; -4- sin ^ cos #. 

y(jc)]7T, a maximum. 

_. N sin x 

15. /(*) = 



i + tan x 

J~(x)~]n, a maximum. 

117. For convenience of reference the following expressions 
for the measures of areas and volumes are introduced. 

Area of triangle = ^ xy, 

Area of rectangle = xy, 

x ~\~ z 
Area of trapezoid = y, 

2 

in which x and z represent the bases and y represents the alti- 
tude. Let 6" = surface, V= volume, x = radius of circle or 
sphere, s = slant height. 

V= 7rx 2 y = volume of cylinder, 

V= = volume of cone, 

3 
V= f irx z = volume of sphere, 

S = 2 irxy = lateral surface of cylinder, 

S = ttxs = lateral surface of cone, 

S = 4 vx 2 = surface of sphere. 



150 Differential Calculus 

PROBLEMS. 

1. Divide a number a into two such parts that their product 
shall be a maximum. 

Let x = one part ; then a — x = the other ; hence, 

/(•*) = (a — x) x, \ 

.-. fix) = a — 2 x = o, .•. x = — • 
Also f"i x ) = — 2, a negative quantity. 

Hence ./"(.*)]„ = — , a maximum ; hence the product of the two 

5" 4 
parts will be greatest when the parts are equal. 

2. Divide a number <? into two factors such that their sum 
shall be a minimum. 

fix) = x H — ; /"'(.#) =i -g = o, .*. x = "fa. A\sof"(x) 

2a 
x 
hence the factors are equal when their sum is a minimum. 



— 3 , which for x = Va is positive ; hence f(x)~\^-, a minimum ; 



3. Divide the number io into two parts such that the square 
of one part multiplied by the cube of the other shall be a 
maximum, 

fix)\4, a maximum, .*. 4 and 6 are the parts. 

4. Required the height at which a light should be placed 
above a table so that the page of an open 

\ book placed at a given horizontal distance 

\ (a) from the light may receive the greatest 

x \ a illumination. 

\ From optics we have the principle that the 

~7A ~7 intensity of the illumination varies directly as the 

~cT^ / sine of the angle of incidence and inversely as 

Fig. 17. th e square of the distance. 



Maxima and Minima 151 

Let i = intensity of the illumination, then from the principle 

and the figure, we have 

sin 

x 

d ax ax 






, (a = intensity at units distance 



d 2 d z (a> + ^f 

from the light). 
Hence, 

.-. (<? 2 + .x 2 ) 2 |# 2 — 2 .x 2 } = o, 
.*. # = — — and ^ = a V— i. 

V2 

As the second value is imaginary, the first value — — only 

V2 

... a 

can satisfy the condition of the problem. Hence —=. is the 

V2 

required height. It frequently happens in the solution of prob- 
lems that the critical value which satisfies the given conditions 
may be detected without analytical examination. Such, for 

instance, as the value —= of x in the above example. 

5. Find the line of shortest length that can be drawn through 
a given point (a, b) and terminate in the rectangular axes to 

which the point is referred. Ans. (a 3 + b*y. 

6. Show that the area of the right triangle formed by a line 
through (a, b) and the co-ordinate axes is a minimum when the 
base of the triangle is 2 a and its altitude is 2 b. 

7. Show that of all rectangles of a given area the square has 
the least perimeter. 



152 



Differential Calculus 



Let m = area, y = altitude, x = base, P = perimeter; 
then P = 2 x -\- 2 y and m = xy, 

2 m 



.'. P =f(x) = 2 x -\- 



X 



2 m 



... /'(*) = 2 - — = o, /. 



a 8 



# 



V#z. 



Hence, 



J = 



;;z 



a- 



.'. the square has the shortest perimeter of all equivalent 
rectangles. 

8. Show that of all triangles of a given perimeter constructed 
on a given base (b) the isosceles triangle has the greatest area. 

/y* I At 1 A 

Here Area = Vj (j — ^) (•$" — j) (r — #) in which s = 

and by condition jc+j4-^ = r, a constant. 

9. A box of maximum con- 
tents is to be made from a rectan- 
gular piece of tin 30" X 14"; 
required the side of the square 
to be cut out of each corner of 
the tin sheet. 

Here v =y(x) = x (14 — 2 x) (30 — 2 x~) 

Ans. x = 3. 

10. Find the maximum rectangle that can be inscribed in a 

given circle. 

Area = 4 xy and a: 2 + j 2 = a 2 ; 

.*. Area = 4 \la 2 x 2 — ^ 4 , 

.*. ./"(#) = ^ 2 ^ 2 — x*. 

Hence, f'(x) = 2 a 2 x — 4 a 3 = o ; 
a 















\ 




1 
1 



Fig. 18. 




X = 



V~: 



= Vtf 2 - x 2 = 






.*. The figure is a square and its 

Area = 4 xy = 2 # 2 . 



Maxima and Minima 153 

11. Find the maximum right cylinder that can be inscribed 
in a sphere of radius a. Ans. Height = § ay/3. 

12. Find the right cylinder of maximum convex surface that 
can be inscribed in a given sphere. Ajis. Height = ay/2. 

13. Find the greatest right cylinder that can be inscribed in 
a given right cone. Ans. Altitude = J altitude of cone. 

14. Find the right cylinder of greatest convex surface that 

can be inscribed in a given right cone. nab 

^Tt ns . o =^ • 

2 

15. Show that the cone of greatest volume and greatest con- 
vex surface that can be inscribed in a sphere of radius (a) has 
f a for its altitude. 

16. Show that the altitude of the cone of least volume that 
circumscribes the sphere is 4 a. 

17. Of all cones of a given slant height show that the one, 
the ratio of whose altitude to the radius of its base is V2, is a 
maximum. 

18. Of all circular sectors of a given perimeter show that the 
one whose arc = twice the radius is greatest. 

Let x = radius ; then, 2 x + arc = p, a constant, and 

Area = \ x (/ — 2 x). 

19. A Norman window, consisting of a semicircle surmounting 
a rectangle, is to be of a given perimeter 
and so constructed that the light ad- 
mitted shall be a maximum ; required, 
the height and breadth of the window. 

With the notation of the figure, we 
have, 

Area = 2 xy -\ , 

2 

and Perimeter = j> = 2jy-{-2x-j- ttx. 




154 Differential Calculus 

7TJC 2 

.'. Area =f(x) = px — 2 x 2 ; 

•'• f( x ) =J> — 4-x — TTX = 0. 

P 



x = 



Hence, y 



4 + 7T 

/ 

4 + 7T 



Therefore the height of rectangle = radius of semicircle. 

We have heretofore, where two variables occurred in the ex- 
pression we desired to investigate for maximum and minimum 
values, substituted for one of the variables its value in terms of 
the other, as derived from given conditions. Thus, in A = 

2 xy -\ above we substituted for y its value in terms of x, as 

2 

determined by the condition p =2 y + 2 x -\- ttx. It frequently 
happens that this substitution may be made more conveniently 
after differentiation. 

Thus, J~( x ) = 2 x y H * 

w/ \ dy 

.'. f (X) = 2 X— f- 2V + VX. 

J dx 



And from 


p=2y-\-2X-\- 7TX 


! have, 


dy 
= 2— — |-2-f-7r. 
dx 




dy 2 + 7r 
' ' dx 2 



dy 
Substituting now this value of — together with that of y 

drawn from the value of p in the value oif(x), we have, 

/ (*) = 2 */ j + 2 ^- - X ~ —J + 7CX 

= p — 4 x — irx, as before. 



Maxima and Minima 155 

20. A person in a boat 3 miles from shore wishes to reach a 
point 5 miles down the 
coast in the shortest time. 
Assuming that he can 
walk 5 miles an hour and 
row only 4 miles an hour, 

at what point must he A q fr 

land? 

From the figure, we have (C being the landing-point), 



x 



CB = V9 + x 2 , CD = 5 - x. 
,\ Total time of rowing and walking is 
V9 -\- x 2 5 — x 



4 5 



! /(*); 



J 4 V9T^ 5 

.•. x = 4. 
i.e., he must land one mile from the point he desires to reach. 

21. What must be the dimensions of a square-based box, open 
at the top, whose volume is 108 cu. in., in order that the mate- 
rial of which it is made may be a minimum ? 

Here S = x 2 -|- ^xy, and V= x*y = 108, 

.\ t S=/(jt:) = ^ 2 + ^. 
.•. x = 6, and j = 3. 

22. From a given quantity of material a circular cylindrical 
cup with open top is to be made ; required its dimensions in 
order that the volume may be a maximum. 

Here V= irx^y and -S = ttx 2 + 2 trxy = c, a constant, 

ex — TTX 3 






x = y 



156 Differential Calculus 

23. Assuming the fact that the area of a segment of a 
parabola is § the rectangle on the ordinate and abscissa, show 
that the greatest parabola that can be cut from a given right 
circular cone is one whose axis = § the slant height of the 
cone. 



Partial and Total Differentiation 157 



CHAPTER XI. 

PARTIAL AND TOTAL DIFFERENTIATION. 

118. Partial Differentials. The partial differ e7itial of a function 
of two or more variables is the differential obtained under the sup- 
position that only one of the variables that e?iters it is changing. 

Thus let u = x 2 -f- xz -f- log y and let d x u, d y u, d z u represent 
the partial differentials of u ; then 

d x u = (2 x -f- z) dx, 

O..U = — > 

y 

d z ii = xdz. 

119. Partial Derivatives. The partial derivative of a function 
of two or more variables is the ratio of the partial differential of 
the function to the differential of the changing variable. 

Thus, in the example of the preceding article, we have, 

du 

—-= 2 x -\- z, 
dx 

du 1 

dy~ y' 

du 

dz 

It will be observed that in writing partial derivatives that the 
subscript in d x u, d y u, d z u, may be omitted as the denominators 
of the derivatives indicate the variable that is supposed to be 
changing. In writing partial differentials it is frequently more 
convenient to use a notation in which the differential of the 



158 Differential Calculus 

changing variable enters. Thus, instead of using d x ti, d y u, d z ii to 
represent the partial differentials, we may use 

die . du du . 

— dx, —ay, — dz. 
ax ay az 

Thus in the example selected we may write, 

du 

— - ax = (2 x 4- z) ax, 
ax 

du , ay 

ay = —, 

ay y 

— az — xaz, 
az 

for the partial differentials. 



EXAMPLES. 

1, If u = x y , show that d x u + d y ii = yx y ~ l dx + x v log xdy. 
Here d x ii = yx y ~ 1 ax and d v ii =x y log xOy, 

.'. d x u + d y u = yx y ~ l dx + x* log #*/y. 

/ n -i . du du 

2. u = log (V* + e y ), show that— + — = i. 

du e* du *> v 

Here — = — — — - and — = 



dx e* -\- e y dy e* -\- e y 

du du e* -f- e y 
dx dy e 00 -\- e y 

i / r~h on i , du du 

3. u = log (x + V^r -|-jr) show that # — +7-7- = 1. 

a 2 v 2 . d& . 3?/ . 2 x . 2 y . 

,,0,0 N , . du du du 

5. z* = log (ar -f- y 5 + » — 3 xyz), show that — — \- — — \- — 

(toe Li_y itz 

= 3 

# +7 4- ^ 



Partial and Total Differentiation 159 

. , du du 

6. u = sin (.ay), show that — — (- — = (# + _y) cos (^y). 

. . , , du du 2 

7. a = log (* +J-), show that^ + - = -. 

8. u = tan -1 , show that x — -f- y—r = o. 

\x -f- j/ ax ay 

120. Euler's Theorem of Homogeneous Functions. 

Let u = ax™/ 1 + bx m 'y nf + ex™" y n " + etc (i) 

be a homogeneous function of x and 7 not involving fractions 

in which 

m -\- n = m' -{- n r = m" + 11" = etc. = p. 

To prove that 

3?/ du 

X T x +y Ty =P "- 

Differentiating (1) partially, we have, after multiplying by 
x, and y, 

x~ = amx m y n + bm x m ' y n ' + cm'Wf" + etc. 

y— = tf/zx m y" + b?i r x m 'y n ' + crf'xr"f" + etc. 
#y 

Hence, by adding, we have, 

* -^ +.y -^ = (/« + n) (ax m y n + fo m y ' + ax^'f" + etc.) =/// . (2 ) . 

The proposition may readily be proved to be true for a 
homogeneous function of any number of variables ; also for 
homogeneous fractions. 

EXAMPLES. 

Verify Euler's Theorem in the following examples by partial 
differentiation. 

1 . u = 4 x*y* -j- 3 x 2 / — 2 xy 4 . 

Here u is a homogeneous function of jc and y of the 5 th degree. 



160 Differential Calculus 

Hence, by (2) Art. 120, 

du du 

x _|_ y — e u. 

dx J dy ■ 
By partial differentiation, 

dit du 

x — — \- y—- = 20 x 6 y z + 1 c; xryr — 10 x \r = zu. 
dx J dy J D y ' ° 

o o du du 

J dx J dy ° 

121. Differentiation by Use of Partial Differentials. The Total 
Differential, or si?nftly the Differential, of a functio?i of several 
variables is equal to the sum of its partial differentials. 

Let // =f(x, y, 2, etc.) ; it is easily seen that 

, du du du . 

du = — dx + —r dy + — dz -\- etc. 
dx dy dz 

in which du in the first member represents the differential of u 
under the supposition that all the variables which enter it are 
changing. For an examination of all the differential forms, 
both of algebraic and transcendental functions, as derived in 
Chapter III., shows that only the first powers of the differen- 
tials of the variables enter the differential of their function. If, 
therefore, we differentiate u =J~(x,y, z, etc.), and collect the 
coefficients of dx, dy, and dz, — these coefficients being usually 
functions of the variables which enter the original function, — 
we may write, 
du = d[f(x, y, z, etc.)] 

= <£ (x, y, z, etc.) dx-\-\\> (x, y, z, etc.) dy -f- w (x, y, z, etc.) dz + etc. 
Now, if we differentiate partially the original functions, we 
obtain, 

du . , , N _ 

-1- dx = (x, y, z, etc.) dx, 
dx y 

du 7 . N _ 

— dy = if, (x, y, z, etc.) dy, 



Partial and Total Differentiation 161 



du 

— dz = o> ( x, y, z, etc) dz. 

dz / 



Hence, adding, 



du du du 

—ax + -rdy H — r dz = <f> (x, y, z, etc.) dx -\- \b (x, y, z, etc. ) ay 

ax ay az 

+ o) (x, y, z, etc.) dz + etc. 

Hence, finally, 

du = -*- dx + -7- dy + -7- dz + etc* 

To illustrate, let us resume the example of § 118, 

u = x 2 -f- xz + log j. 

Here, du = —dx -\ — — dV -| — - tfs 

dx dy dz 

=■ (2 x -\- z) dx -\ (- xdz. 

y 

Obviously the example above may be differentiated by the 
rules deduced in Chapter III. ; and, as a matter of fact, those 
rules are sufficient to enable us to differentiate any algebraic or 
transcendental function, whether explicit or implicit. In cer- 
tain forms of expressions, however, it will be found more con- 
venient to adopt the process here explained. 

Cor. I. If u =J'(x,y) = c, a constant, be an implicit func- 
tion of y, then 

du du 

du = — - dx -f- -7- dy = o. 
dx dy 

du 

u dy dx 

hence, -^ = — -_ 

ax du 

dy 

i.e., the first derivative of an implicit function is minus the ratio 
of its partial derivatives. 



162 Differential Calculus 

Let 



7/ == a 2 y* + l?x? = a 2 lr ; then 


dH 72 

— - = 2 /rx, 
awe 


du 

— - = 2 rtry. 


6// 




*/>> dx 
dx du 


2~~' 


Ty 





Hence, 



a result previously obtained by direct differentiation (see Ex. 8, 
p. 26). This method will also be found convenient in differenti- 
ating many cases of implicit functions. 

122. Total Derivative. The total derivative of a function of 
several variables is the ratio of the total differential of the 
function to the differential of the independent variable that 
enters it. 

Thus, if ?/ =J~(x,y, z } etc.), then (§ 121), 

, du . du du . , . 

d u = — dx -\- — dy -\ — - dz -\- etc (a) 

dx dy dz 

Now, if x = cj> (e'), y = \p (7'), z = w (v), etc. ; then ti is indi- 
rectly a function of v through x, y, and z. 

Dividing both members of {a) by dv, we have, 

du __ du % d x , chu % dy . du ^ d_z , , ,,v 

cfo c/j: cfa> c/3; cfo cfe cfo 

To illustrate, let 7/ = x z — xy -f- log z, in which .*: = v 2 , y = sin v, 
and s = ^ v ; then 

du _ //.r dzv dy du 1 

— - = 3 ^ — y\ — = 2 v; -=- = — x ; -7- = cos z/ ; — = - , 
dx dv dy dv dz z 

dz 

dv 



Substituting in the formula (#), we have 

du 

dv 



du e v 

— = 2 (3^ — j^)z/ — # cos z/ + — = 67/— 2 7'sinz' — v 2 cosv-{- 1 



Partial and Total Differentiation 163 

for the total derivative. The same result may be obtained by- 
substituting the values of x, y, and z in the value of u and then 
differentiating; thus, 

u = z> & — v 2 - sin v + log e v . 

Hence, -7=6^-2^ sin v — v 1 cos v + 1 , as before. 
dv 

Cor. 1. If u =J~(x,y, z) in which y = <f>(x), z= \p(x)\ then 

du du du dy du dz 
dx dx dy dx dz dx 

Cor. 2. If u =/(x,y), in which x= <f> (v) and y= \p (ji) ; then, 

du . du 
du = — dx 4- — dy, 
dx dy 

du du dx du dy 
dv dx dv dy dv 

Cor. 3. If u —f(x, y) and y = <f> (x) ; then 
du du du dy 
dx dx dy dx 

Cor. 4. If u =f(y) and y — <f> (x), then, 

du du dy 
dx dy dx 



EXAMPLES. 

By aid of partial differentials differentiate the following, and 
verify the results by direct differentiation : 

du = yx y ~ 1 dx + x y log xdy. 

du = yzdx + xzdy -+■ xydz. 

ydx — xdy 
du =< — -. 

4. u =^ 5 +logsinj-f-coss, du = 5 x 4 dx + cot^ dy — sin zdz. 



1. 


u = x y , 


2. 


u = xyz, 


3. 


X 

u = -, 

y 



1G4 Differential Calculus 



5. u = x losy , 



{ * y ) 



6. u — a x e y , du = a x e v jlog adx -+- dy\. 

, y T xdy — ydx 

7. u = tan" 1 ^- , du = ^ , % . 

X xr -+■ jr 

, - .%• 7 ydx — xdy 

8. # = log tan v - , du = 



y (^ 2 + /)tan- 1 - 

y 

Write the first derivative of each of the following implicit 
functions by use of partial derivatives, and verify results by the 
direct process : 



9. a 2 f - b 2 x 2 = - a 2 b 2 . 
Let u = a 2 y 2 - Px 2 + a 2 P. 

Then, § 121, Cor. i, we have, 

du 

dy dx — 2 fix ftx bx 









dx du 2 a 2 y a 2 y a ^ _ a * 

dy 
Directly : 2 a 2 ydy — 2 b 2 xdx = o, 

dy b 2 x 
dx a 2 y 

1A „ ~ fy y(j — x\ogy) 

10. x y — f = o, — = — i 

dx x{x — y log x) 

ye^y dy my 

11. — = a, 



x m dx x (1 + ny) 

12. sm(xy) + tan (#1;) = m, — = 

</y a - y cos *y 

13. sin (jev) — tf.x = o, — = ■ — 

v ' #:*•.*: cos jey 



Partial and Total Differentiation 165 

Find the total derivative of each of the following : 

14. u = x z -\- y 2 — xz, in which x = v 2 , y = sin v, z = log 7/. 

Here — - = — — T--\--j--r-\ — r~r m See Art. 122. 

dv ax dv ay dv dz dv 

du du du 

Also _ = 3^ + 2, — = 2 y, — = — x, 

ax ay dz 

dx dy dz 1 

— = 2v, -j- = cos v, — = - ; 
dv dv dv v 

. du , „ N # 

hence, — = (;r-0)2z/-|- 2v cos v 

dv vo y ^ z/ 

— (3 ^ 4 — l°g v ) 2 v ~r~ 2 sni & cos v — v 
= 6 z/ — 2P log z> + sin 2 v — v. 

To obtain this result directly we substitute the values of .r, y, 
and 2 in the value of u and obtain 

u = v G -{- sin 2 7' — z> 2 log v. 

Hence, — - = 6 v 5 -\- 2 sin v cos z/ — [ z> 2 — h 2 7/ log v ) 
dv \ v ) 

= 6 z> 5 + sin 2 z> — z> — 2 z/ log z/, 
as before. 

15. ^ = y 2 -\- z* -\- sin xy, y = x s , z = tan ^. 

^ du du dy du dz _ _ 

Here — = — + -3- -^ + ~r ^- • See § 122, Cor. i. 
# x ^ dy dx dz dx 

.-. — = 6 x 5 -f- 4 (tan 3 .»: sec 2 ^ + ^ 3 sin ,# 4 ). 

16. ?/ = j 3 -f- sy + z 2 , y = **, 2 = sin at. 

*/« du dy du dz ^ ' „ 

Here — = — ^+— — . See § 122, Cor. 2. 

^x dy dx dz dx 

du „ . s 

.•. — = 7. <r x + ^ (sin jp -j- cos #) + sin 2 #. 



166 Differential Calculus 



y 

17. u = tan -1 - , v 2 = 2 A*r. 

du du du dy „ 

Here — = — + — -f . See § 122, Cor. 3. 

*/.* ax ay ax 



du *Jpx 



dx (^_|_ 2 px)\f2 % 

. x . du 1 

18. 2/ = log - , y = sin jc. — - = cot x. 

y dx x 

19. u = tan^y 2 , j = log x. 

du du dy 
Here —= — -=-, See § 122, Cor. 4, 

dx dy dx 

du 2 log x sec 2 (log jc) 2 
dx x 

z du (f' €** 

20. u = sin - , y = x 2 , z = e*,. — = (x — 2) -z cos -^ 

jy </# ' ar ar 

,^-y _ du 2 e* x 

21. » = tan x t z = e* f y = e x . — - = 



2 +-J *£c ^"+ 1 

1 / - du -i 

22. # = sin -1 (j — z), y = 3 x, z = 4 jr. — = 



dx Vi — J^ 2 



jv k jc v •# ^/z/ 



SUCCESSIVE PARTIAL DIFFERENTIATION. 
123. Successive Partial Differentials and Derivatives. 
In general, if u =f(x,y), then 

— dx = <f> (x, y) dx (a) 

du 
and -j-dy = if/(x,y)dy (b) 



Partial and Total Differentiation. 167 

i.e., the partial differentials of a function of x and y are, in general, 
functions of x and y. Hence, differentiating again, with respect 
to either x or y, regarded as equicrescent, we obtain a second par- 
tial differential. Thus from (a) we have, 

d (du \ d 2 u 

—-[—r-dxXdx = ~=-s dxr = cb-. (x, y) dx 2 , 
dx \dx ) dxr N J 

Ty {£ dx ) dy = !jk dxdy = ^ X ' J) dxdy ; 

and from (&), 

d (du , \ , d 2 u 
dy\Ty dy ) dy = If dy = ^ <* y) df > 

d (du . \ . bhidydx . N 

- ^- j dx = -j-^ = ^ (x, y) dydx. 

As the second partial differentials are also, in general, func- 
tions of x and y, we may again differentiate and obtain a set of 
third partial differentials, and so on. Thus, from the notation 
adopted above, we have 

!My d ^ dy 

for the symbol of the result obtained by three successive partial 
differentiations, huo of these being with respect to x, and one 
with respect to y. Hence, 

B 3 u 
dx 2 dy 

is a symbol of the third partial derivative obtained by the same 
process. Other symbols of a third partial derivative are obvi- 
ously, 

b z u d 3 y d 3 u 

dx 3 dy s dxdy 2 

and similarly for other derivatives. 



168 Differential Calculus 

124. To prove, 

d fdu\ d /du 
dx \dyj dy \dx 

i.e., that 



dxdy dydx 
Let u = f(x, y) ; thus, regarding y as a constant, we have, 

Au f(x + Ax, y) —f(x, y) 

Ax Ax 

Ex. 10, p. 52. Now, regarding x as constant, we have, 

\Ax/ = /( x + Ax, jy + Aj) —f(x,y + Ay) — /fo + A.r,j) +/(*, j) 
A_y AjAjc 

(0 

Reversing the above order, we have, 

Au _f(x, y + Ay) —f(x, y) 
Ay Ay 

and 

\Ay/ _ f(x + Ax,y+Ay) —f(x + Ax,y) —f{x,y + Ay)+f(x,y) 

Ax AxAy 

(*) 

Hence, J&l = J&I . 

Ax Ay 

Passing to limits, we have, 

dx dy 

d 2 u d*u 

i.e. ■ = * 

dxdy dydx 



Partial and Total Differentiation. 169 

Cor. Similarly we may prove, 



dx?dy dydx 2 
d n u d n u 



dx n ~ s dy s dy s dx n ~ 3 

i.e., whatever the number of differentiations the order of differ- 
entiatiofi is immaterial. 







EXAMPLES. 


1. 


u = x h y — 


d 2 u d 2 u 
' dxdy dydx 


w 

Al 


e have, 
so, 


du . 

— = x* — cos y ; 

dy 

bhc 
dxdy 

du 

5 = *** 



Hence, 



d 2 // 

d 2 z/ d 2 // 



d 2 /*; d 2 u 



2. u = x log (ay — i) ; prove that 



dxdy dydx 



3. * = xy (x + /) ; show that -JL = ^JL. 

d 2 u d 2 u Bu 

4. « = (x*+f)*; showthat 3*— + 3/^ + ^ = 0. 

5. * = cos (* + jr) s show that Jj = ±jL. 



170 Differential Calculus 

125. To find the successive total differentials of a function of two 
independent variables. 

Let u = f(x,y); thus, (§ 121), 

_ du du . 

du = — dx -\ — - dy. 
dx dy 

Differentiating, remembering that — and — are, in general, 

functions of x and y, and that x and y, being independent, may 
be regarded as equicrescent, we have, 

,_ d 2 u _ „ d 2 u . _ d 2 u _ , d 2 u _ . 

d*u = -r-= tf.r J -f - — - ardfy -f- -r~r dydx + -7-5 #r 
dxr dxdy dydx dy 

^ u v 2 . ^ 2 u , 6> 2 // 

' = *?** + ' *&** + &&■ 

Similarly, we find, 

-o d 3 « d 3 // , „ , d 3 u _ , _ 3 3 z/ „ 

^jc 3 ^ ° /£eVy ^ ° dk/Zy 8 J dy 3 J 

and so on. By observing the analogy between the exponents of 

du, d 2 u, du 3 , . . . 

and those of the development of 

(x + a), (x + a) 2 , (x + a) 3 , . . . 

we are enabled to write the value of d n u. 

The student may apply this process to any example. 



Direction of Curvature Points of Inflexion 171 



CHAPTER XII. 

DIRECTION OF CURVATURE. POINTS OF 

INFLEXION. 



CARTESIAN CURVES. 

126. A curve is concave upward or convex upward at a point 
according as the tangent at the point lies below or above the 
curve. 





Fig. 22, a. 



Fig. 22, b. 



Thus, Fig. (a), the curves J/TVand M' N' are concave upward ; 
and, Fig. (b), the curves ST, S' T' are convex upward at the 
points T, P'. 

127. Investigation for Direction of Curvature. 

Let y =f(x) 

be the equation of any curve ; then 

is the slope of the tangent to the curve at the point (x,y). § 19. 



172 



Differential Calculus 



It is readily seen from Fig. (a), by examining either of the 

dv 
curves MN or M' ' N 1 , that — increases as x increases ; hence 



dx 



dy 
dx 



\_=f r (x)\ is an increasing function of x\ hence, § 28, Cor., 



dx 2 



> 0. 



On the other hand, the slopes of the tangents to the curves ST, 

dy 
S' T' ', Fig. (d), are decreasing functions of x, i.e., — \j=/ r (x)\ 

decreases as x increases ; hence, at such points as P, P r 

J 2 y 



dx 2 



<o. 



Hence, in general, a curve y —f(x) is concave upward or con- 
vex upward at the point (x, y) according as -7-^ > or < than o. 

128.* Point of Inflexion. The point at which the direction 

of the curvature changes is called a 
point of inflexion. 

Such a point is P, Fig. 23. Since 
the curvature changes from concavity 
to convexity upward, or from con- 
vexity to concavity upward at a point 

of inflexion -— changes sign from -f- 

to — , or from — to + ; hence at the 
point 




Fig. 23. 



dx" 



or 00. 



To determine, therefore, whether any given curve has a point 
of inflexion we obtain the second derivative from its equation, 

* Sluze, in 1659, pointed out a general method for determining points of inflexion by 
reducing it to a question of maxima and minima, viz., by investigating for a maximum or 
minimum intercept made by a tangent on any axis from a fixed point. 



Direction of Curvature Points of Inflexion 173 

and equate the result to zero or infinity. The roots of this 

equation are the critical values. If for values of x a little less 

d 2 y 
and a little greater than any one of these critical values -=-^ 

changes sign, then for that critical value there is a point of 
inflexion. 

To illustrate, let us examine the equation 

6 y = c (x — of 
for a point of inflexion. 

Here — — „ = c (x — a) = o, 

ax 1 / 

.'. x = a is a critical value ; and as -^ obviously changes sign 

as x passes through the value a the curve has a point of in- 
flexion at (a, o). 

EXAMPLES. 

1. Determine the direction of curvature of the parabola 

y 1 = 2.pX. 

Here ^ = -t. 

dx 2 f 

d 2 y 
For negative values of y, -^— > o ; for positive values of y, 

d 2 y 

-z-j < o ; hence the parabola is concave upward below the 

^r-axis and convex upward above the ^c-axis. 

2. Show that the hyperbola xy = m is concave upward in 
the first angle and convex upward in the third angle. 

d 2 y 2 m 
Here 

3. Show that 3 a 2 y — x z + 3 ax 2 — 6 a z = o has a point of 
inflexion at (a, f a), and that the curve is convex upward on the 
left of this point and concave upward on the right. 



174 



Differential Calculus 



4. Examine the witch y = 



Sa* 



x 2 + 4a 2 



for points of inflexion. 



Points of inflexion ( —=a, -ai 
V3 2 



2 3 
—p=-Ci, —a 

^3 2 



5. Examine the logarithmic curve x = log y for direction of 
curvature, and show that it has no point of inflexion. 

6. Show that the " curve of sines " y = sin x has an infinite 
number of points of inflexion, and that the ordinate of each is o. 



7. Show that the curve - = 
flexion at x = - . 



a t 2 a — x 



has a point of in- 



8. Find the abscissa of the point of inflexion in the curve 



x 



xy = a 2 log - 



Am. x = ae*. 



POLAR CURVES. 



129. A polar curve is convex or concave to the pole at a 
point according as the tangent to the curve at the point does, 
or does not, lie on the same side of the curve as the pole. 





Fig. 24. 



Thus, Figs. 24, the curve MN is convex, and ST is concave 
to the pole O. 



Direction of Curvature Points of Inflexion 175 

130. Investigation for Direction of Curvature. 

Let r=f(0) 

be the equation of either of the curves MN, ST, Figs. 24, 
— MN and ST being any two curves referred to polar co- 
ordinates. Let PB, P' B' , be two tangents drawn at any two 
points P, P '; and let OB, OB' be perpendiculars let fall from 
the pole on these tangents. From Fig. 24 (a) we see that as 
r (OP) increases/ (OB) decreases, and from Fig. 24 (b) that 
as r (OP) increases/ (OB) increases ; hence, in either case, 

p = F(r), 

and since p is a decreasing function of r when the curve is 
convex and an increasing function of r when it is concave, 

we have 

dp 

— - < o or > o, 

dr 

according as the curve is convex or concave to the pole. 

From § 78, we have 

_ t 3 - 

I . . /drV' 



\/"+(5) 



hence to investigate any curve r =/($) for direction of curva- 

dv 
ture at a given point we first obtain — from the equation of 

. du 

the curve and substitute the square of the value found in the 

above value oip, the first derivative of the resulting expression, 

( — ), in which the coordinates of the given point have been 

substituted, will determine by its sign whether the curve is 
concave or convex to the pole at the given point. 

Thus let us examine the spiral of Archimedes r = aO for 
the direction of curvature at the point (r, 6). 



176 Differential Calculus 



Here -- Q = a, 

hence, p = 



Vr> + a 2 



dp r s -j- 2 # V 
^ ~ (V s + « 2 ) f ' 

which is positive for all positive values of r ; hence the curve 
is everywhere concave to its pole. 

131. Point of Inflexion. Since the tangent to a curve at a 
point of inflexion crosses the curve (see Fig. 23), the curve is 

convex to the pole on one side of the point, and concave to the 

dp . 

pole on the other side ; hence, — changes sign from — to +, 

dr 

or from + to — ; hence, at the point, 

dp 

— - = o or 00 . 
dr 

To examine a polar curve, therefore, for a point of inflexion 

dp . 

we obtain -4- as in the preceding article, and ascertain what 
dr 

values of r, if any, reduces this derivative to o or 00 . If such 

dp 
values of r exist we then ascertain whether or not — changes 

dr 

sign as r passes through this value ; if it does, the critical value 
corresponds to a point of inflexion. Thus in the Spiral of 

Archimedes, discussed in the preceding article, we find by 

dp 
equating — to o that 



r = o and r = a V— 2 

are critical values ; and by equating it to 00 that r = a V— 1 
is another. As two of these values are imaginary and the other 
(r = o) corresponds to the starting-point of the curve there is 
no point of inflexion. 



Direction of Curvature Points of Inflexion 177 

EXAMPLES. 

1. Examine the lituus, r 2 9 = a, for direction of curvature and 
for points of inflexion. 

Here 

= 



dO 2 r(P 2 a 

2 ar 



V4 a 2 + r 4 



dp 8 a 3 — 2 ar* 2 a (4. a 2 — r 4 ) 

Hence, -±- = = = /- = o. 

dr (4 a 2 + r 4 y (4 a 2 + r 4 ) 5 

.*. r 4 — 4 <2 2 = o, 

.'. r = \l2a. 

Hence, when r < \l2~a, — - > o, .-. curve is concave to the 

dr 

pole ; and when r > VT^, — < o, .'. curve is convex to the 

ar 

dp 

pole. Since — changes sign as r passes through the value 

V2 a, that value of r corresponds to a point of inflexion. 

2. Show that the logarithmic spiral r = a 9 has no point of 
inflexion, and that it is everywhere concave to the pole. 

3. Show that the curve rO = a esc has a point of inflexion 

. 2 a 
when r = 

4. Show that the curve r6 m = a has a point of inflexion when 
= Vm(i — m). 

5. Show that points of inflexion on polar curves can be 

d 2 u 1 

determined by aid of the equation u + —- = o where u = - . 

6. By aid of the equation given in Ex. 5, show that the curve 

f *J /J 

r=(r—a)6 2 has a point of inflexion at (— , V3) ; also at 



178 



Differential Calculus 



CHAPTER XIII. 

CURVATURE. CIRCLE AND RADIUS OF CUR- 
VATURE. EVOLUTE AND INVOLUTE. 



History. — Huygens, in the third chapter of his Horologium Oscillato- 
rium (1673), defines evolutes and involutes, proves some of their more ele- 
mentary properties, and illustrates his method by finding the evolutes of 
the cycloid and the parabola. 

132. The Measure of the Curvature, or more simply, The Curva- 
ture of a curve, is the ratio of the rate of change of its direction 
to the rate of change of its length. 

Let a be the angle which 
the tangent to the curve 
MN at P, Fig. 25, makes 
with the X-axis and let NP 
= s ; then, since the direc- 
X tion of a curve at a point 
is the same as that of the 
tangent, we have da. = rate 
of change of direction of 
the curve, and ds = rate of 
Hence, by definition, 




Fig. 25. 



change of the length of the curve. 

c = — 
~ ds 

in which c represents the curvature of MN2I the point P. 

To show how this ratio measures the curvature let us sup- 
pose the curve MN to be generated by the point P moving 
with any velocity v and carrying its tangent along with it as it 



Curvature Evolute and Involute 179 

moves. . Let us further suppose the curve SL (tangent to 
MN and PT at P) to be generated by a coincident point P 
moving with the same velocity v. Then, by definition, the 
curvature of SL at P is 

f da 

c = w 

in which da is the rate of change of the direction of PT, the 
tangent to SL, at P, and ds is the rate of change of the length 
SP = /. But since the generating points move with the same 
velocity, we have 

v = ds = ds r §17; 

, C da 

hence, -, = -yj 

c da 

i.e., the curvature of two curves at any two points are to each 
other as the rates of change of their direction. 

For example, let da = 30 a second and da = 6o° a second ; 
then 

L - 3^° - I 

C OO 2 

or c' = 2 c, 

i.e., the curvature of one curve is twice that of the other. 

133. Circle of Curvature. Radius of Curvature. 

The circle tangent to a curve at a point, and having the same 

curvature as the curve at the point, is called the Circle of Curvature, 

for that point. The Radius of Curvature is the radius of the cii'de 

of curvature, and the Center of Curvature is the center of this 

circle. 

Thus, Fig. 26, if the circle CPP r has the same curvature as 
the curve NM at the point P, CPP r is the circle of curvature 
for that point ; OP, the radius of CPP' , is the radius of curva- 
ture, and the center O, the center of curvature. 

It is obvious that the circle of curvature and radius of curva- 



180 



Differential Calculus 



ture vary from point to point as the curvature of the curve 
changes. 

Since, by definition, the circle of curvature CPP' and the 
curve JVM have the same curvature at P, we have, § 132, 

da 






ds 



for the curvature of the circle at P. But if we suppose the 
length of the arc PP'(= ds) to be the velocity with which the 




Fig. 26. 

generating point passes through P, then POP' = a — a repre- 
sents the angular velocity of the tangent, i.e., the rate of change 
of its direction. Hence, 

POP' = da. 

Let OP = p, then from the circle, 

arc PP' 



POP' = 



i.e. 



da = 



Hence, 



ds 

P 

ds 1 



da C 
i.e., the radius of curvature is the ?'eciprocal of the curvature. 



Curvature E volute and Involute 181 

ds r 
Cor. i . If p — -j-j be the radius of curvature at any other 

da J 

point of the same curve, or at some point of another curve, 

ds 
da 



we have ^ = 

P 

da' 



ds" 



da 

p ds' c 
i.e., — = — — = — . 

p aa c 

ds 

Hence, the curvatures of a curve at any two points are in- 
versely as the radii of curvature at the -points. 

134.* Expressions for the Radius of Curvature, 
i. In Terms of Rectangular Coordinates. 

Remembering that ds = {do? + dy 2 y, § 18, and that 

dy 
a = tan -1 —- , § 19, we have 
dx 



ds 
P = da 



(dx 2 + d/y 

i dy 

^/tan- 1 ^ 

dx 

(dx 2 + dy 2 f 
d 2 y 
dx 



1 + W 

dx 2 



* Equation (i) § 134, was first given by John Bernoulli in 1701. 



182 Differential Calculus 



i+(grj' 

hence, after reduction p = - ^ — — — (i) 

d 2 y v 7 

dj? 

If j is taken as the equicrescent variable, we have (Ex. 9, 

P ' I04) ' $ wf 

Cor. If in (1) - r -= = o, p = 00 , .". £ = — = o : i.e., at a point 
# jr p 

of inflexion the curvature is zero. See § 128. 

2. In Terms of Polar Coordinates. 

1 r df) r 

In this case ds = (r 2 d0 2 + dr 2 Y, §76; tan <£ = — - = ■—. 

dr dr 

r — 

.•. <f> = tan -1 —- : a = 6 + <£• See § 77. */0 

Hence, p = — 

da 

_ (fd& + dr 2 ) h (W0 2 + </r*)* 



d(0 + <t>) dd + dcf> 

dO + </tan- 1 4- 
dr 

dO 

(r 2 dP + dr 2 f 

dr . d 2 r 

— dr — r 

dO d$ 



d0 + 



dr\ 2 
dOj 



r 2 
1 + 



/drV 

\de) 



Curvature Evolute and Involute 183 

Hence, reducing, we have 

P= < Wi (X) 

2 A/A* rfV 

If r is taken as the equicrescent variable, we have (Ex. 8, 
P- io 3)> 

r +r U 

d~0 2 /d$\ 3 dd ' ' ' ' W 
dr 2 \drj dr 

Cor. Since p = oo at a point of inflection (§ 134, Cor.), we 

/dr\ 2 d 2 r 
have, from (1), r 2 + 2 1-^1 — ^-ttw = o as a necessary condition 

for such a point. 

135. At a point of Maximum Curvature the circle of curvature 
lies e?itirely within the curve. 

For on either side of the point the curvature of the curve is 
less than at the point, while the curvature of the circle of curva- 
ture is the same on either side of the point of tangency as it is 
at that point. Hence, the circle of curvature lies entirely within 
the curve. 

Similarly, we may show that at a point of minimum curvature 
the circle of curvature lies entirely without the curve. 

Cor. Since, in general, the curvature of a curve at a given 
point is less than at the preceding consecutive point and greater 
than at the following consecutive point, or vice versa, we have, 
as a general proposition, that the circle of curvature crosses its 
curve. 



184 Differential Calculus 



EXAMPLES. 



1. Show that the curvature of the circle x 2 + y 2 = a 2 is con- 
stant, and find its radius of curvature for the point (x, y). 



Here, c = 



d^y 

i dx 2 



' \-*m 



From the equation x 2 + y 2 = a, we obtain 



Hence, 



dy 


X X 


dx 
d 2 y _ 


y \la 2 -x 2 
a 2 _ a 2 


dx 2 


y VO 2 - x 2 ) 3 
a 2 




(a 2 - x 2 f i 




( #2 ) § a 



a" — x* 



Hence, the curvature of the circle is constant, as previously 
assumed, and is the reciprocal of its radius. 

Again, p = — • .*. p = a, 

i.e., the radius of the circle is its radius of curvature. 



2. Find the radius of curvature of the logarithmic spiral 

r = ^ e . 

( dr\ 2 J f 



Here, p = — 



dr\ 2 d 2 r 



Curvature Evolute and Involute 185 

From the equation we have 

dr . d 2 r „ „ 

d0~ ' dd 2 ~ ' 



e* 9 + 2 a 2 e 2ad - a 2 e 2a9 
= r Vi + d 2 . 
Find the radius of curvature of each of the following : 

2 \J(x + tf) 3 



3. y = 4 ##. p = 



4. 


jc 2 y 2 

-2 + ^2 = I - 


5. 


r 2 = # 2 cos 2 0. 


6. 


r = a(i — cos#). 






p_ ^ 4 



P 3 r 



p = f V2 ar. 

7. y = -(<?«+ e~"). p = - (e* + <T 5 ) 2 . 

2 v 4 v 7 

8. x = a vers -1 V2 ay — jj/ 2 . p = 2 V2 #y. 

9. 2 xy = dr. p = — — • 

a 2 

2 2 2 3/ 

10. x 3 -\- y 3 = a 3 . p = $yaxy. 



y 






11. ^ a =sec-« p = <z sec — ■ 

a a 

J J 

12. r=«sec -• p= 2«sec -• 

2 2 

136. Evolute. 77z<? Evolute of a curve is the locus of the centers 
of the circles of curvature of the curve. 

137. Involute. The Involute is the curve whose centers of 
curvature form the Evolute. 

Thus, if the curve N' (f S is the locus of the center of curva- 
ture of the curve NPM, N' C? S is the evolute of NPM, and 
NPM is the Involute. See Fig. 27. 



186 



Differential Calculus 



138. Equation of the Evolute. 

Let_y =/{x) be the equation of involute NPM (Fig. 27), and 
let (f be the center of curvature corresponding to any point P 
on the involute. Let (x, y), (x',y'), be the coordinates of P 
and (7, respectively. It is required to determine (1st) the 




Fig. 27. 



values of x' andj/ in terms of x and y, and (2d) to determine 
from these values in conjunction with the equation of the invo- 
lute, y =f(x), a general relation between x r and y\ i.e., to 
determine the equation of the evolute. 

1. To determine the coordinates of the center of curvature (x f ,y'y 

Let <yp = p, and draw PD ± OB. Since C^'isJ-to the 
tangent PT, we have DC? P = a. We have, therefore, 

OB = OC — PD = OC - p sin a. 
But OB = x\ OC = x, sin a = -£ = 



, (§ 18). 



Substituting these values together, with that of p, as given in 
§ 134, we have, 

. \ \dxj \ dx , . 

x = # — \„ (1) 



d 2 y 
dx 2 



Curvature Evolute and Involute 187 

Again, OB = PC -f- D 

= PC + p COS a. 

But <9'.Z? = /, ^C = j/, cosa = ^ = 



dy^ 2 



\dx j 

■*■ /=y+ —d^r • ■ (2) 



dx* 

2. To determine the equation of the evolute. 
If we now combine with (i) and (2) the equation of the 
involute, 

y-A x ) (3) 

eliminating the variable coordinates x and y of the involute, 
we shall have a resulting equation in x r and y r . The equation 
thus obtained is obviously that of the evolute. 

To illustrate, let us find the equation of the evolute of the 
parabola, y 1 = 4 ax. 

Differentiating we obtain 



*! = \ ft , d% y = 1 1 /Z 



Hence, #' = # + 



a I A a 



v£ 



, # — 2 # 

i.e., # = 3 # -}- 2 tf , # = 



Also, 7 = j — 



m 



f 



i.e., /=-—-, y = - (4 a 2 /)*. 

4 # 



188 



Differential Calculus 



Substituting these values in the equation of the parabola, we 
have, 



x — 2 a 



i.e., 



27 ay" 1 — 4 (ptf — 2 df 




is the equation of the evolute. 

The semi-cubic parabola CDE is the locus of this equation, 

the branch DE being the 
E evolute of OB, and DC the 

evolute of OA. 

Since y r — o, gives x' = 
2 a, the vertex D is at a 
distance from the origin equal 
to the semi-latus rectum of 
the parabola y 2 = 4 ax. 

An inspection of the figure 
shows that OD is also the 
radius of curvature of the 
curve at o, and since it is 
evidently a minimum, the curvature of the parabola is a max- 
imum at the origin. 

139. Properties of the Evolute. 

1 . A tangent to the evolute is twrmal to the mvolnte. 

Let N'S, Fig. 
29, be the evolute 
and NM the in- 
volute. Let PA 
be a tangent at 
any point P (V, 
y r ) of the evolute, 
and TA a tangent 
to the involute at 
the point A(x, y) where PA cuts the curve. We wish to show 
that PA is J_ to TA ; i.e., that 




T 
Fig. 29. 



-X 



Curvature E volute and Involute 189 



TV 



r I 

a = \- a. 

2 

From § 138, we have 

/ = y + p cos a > 

x' = x — p sin a. 

Differentiating these equations, we have 

dy' = dy — p sin ada -f- cos adp, 

dx r = dx — p COS ada — sin a//p. 

ds dy ds dx 

But p sin cu/a = — — r da= dy, and p cos ada = — — — da = ^r, 
da dfr da ds 

§§ 133, 18; hence, 

dy' = cos a</p (a) 

dx' = — sin adp (J?) 

therefore, by division, 

dy 

——, = — cot a ; 
dx 

i.e., § 19, tan a' = tan j — f- a ) ; 

/ 7r 
.*. a = |- tf, 

2 

and the tangent to the evolute is normal to the involute. 

2 . 77^ difference between two 1'cidii of curvature is equal to the 
length of the a?r of the evolute between the corresponding centers of 
curvature. 

Thus, Fig. 29, we wish to prove that 

P'A'- PA = PP'. 
Squaring and adding equations (a) and (//), we have 
dy' 2 + dx' 2 = dp 2 (cos 2 a + sin 2 a) ; 
hence, ds' 2 = dp 2 ; 

.*. ds' = dp, 

i.e., the rate of change of JVP(= s') is equal to the rate of change 



190 Differential Calculus 

of PA (= p) ; but since any interval of time may be regarded as 
a unit of time, we have ds' = PP' and dp = P' A' — PA ; hence, 

P'A'-PA=PP'. 

140. The properties of the evolute demonstrated in the pre- 
ceding article afford a method of describing the involute 
mechanically when the evolute is given. Take a string of any 
length, and wind it around the evolute JV'S, Fig. 29, one end of 
the string being at JV. If we place the point of a pencil in a 
loop of the string formed at N, and unwind the string, the pen- 
cil-point will describe the involute. For the string is always 
tangent to the evolute N' S and normal to the involute JVM; 
also, the difference between the unwound lengths in any two 
positions, say P' A' , PA, is evidently equal to the arc of the 
evolute {PP') between the points of tangency. 

Since the curve JV'S is usually of indefinite length, there may 
be an infinite number of involutes of which it is the evolute ; 
there can, however, be only one evolute of which it is the 
involute. 

EXAMPLES. 

1. Find the evolute of the circle, x 2 -f- y 2 = a 2 . 

dy x d 2 y a 2 

dx y dx 2 y z 



Here, 



x*\ x 

y 2 J y 

Hence, x'= x — = x — x = o. 

a 2 

7 

X 2 
f 

y = y — =y - y = o. 

7 

Hence, the evolute is a point. 



/ 



Curvature Evolute and Involute 191 

X 2 V 2 

2. Find the evolute of the ellipse, — + -^ — i. 

cr tr 

dy fix d 2 y b 4 

dx a 2 y dx 2 a 2 y z 

Hence, *>= <* ~J>* , /= - <*-?* 

Substituting the values of # and _>> drawn from these values 

of x' and y in the equation of the ellipse and reducing, we 

have 

x'l y§ _ (a 2 - F)% 



7 2. ' 2 2/2 

#3 #3 tf3#3 

for the equation of the evolute. 

3. The hyperbola,- -- = i. ^ = ^ — ri ^— 

ar tr fa a* a^b* 

4. The tractrix, x = a log V# 2 — JP 2 , has for its 

evolute the catenary, y = - (^ — e a ). 



2 







5. The cycloid, x = a vers 1 \J2 ay — y 2 . 



x r = a vers" 



•i(_j)_V2 a y-y. 



6. The hypocycloid, (-) 3 + (-) = i. 



a/+y\ / *'- / y =2 



7. The equilateral hyperbola, 2 jy; = <z 2 . 

1= 2. 



/ *'+y y /x'-yy 



192 Differential Calculus 



CHAPTER XIV. 

CONTACT OF CURVES. ENVELOPES . 

History. — Envelopes may be said to have originated with the investi- 
gations of Huygens on evolutes and those of Tschirnhausen (i63i-i7o8)on 
caustics. 

Leibnitz laid the foundation of the theory in a memoir written in 1692. 

141. Orders of Contact. Let y = <j> (x) and y = ^ (x) be the 
equations of two curves, and let x = a be the abscissa of a 
common point ; then 

cf> (a) = if/ (a), 

i.e., their corresponding ordinates are equal. If we suppose, 
moreover, that the curves touch each other, we have 

4>'(a) = tf(a), 

i.e., the curves have a common tangent at the point. When 
these conditions are fulfilled for any two curves they are said 
to have a Contact of the First Order. 

If (f> (a) = iff (a), 4>'(a) — ^'00* an d, also, 

4>"(a) = «A», 

then the curves not only touch each, but their curvature at the 

common point is the same since A— -1 is a function of the 

first and second derivatives only. See § (134), 1. Under these 
conditions the curves are said to have a Contact of the Second 
Order. 

If cf>(a) = if/ (a), 4>'(a) = <//(», <f>"(a) = $"(a), and also, 



Contact of Curves Envelopes 



193 



the curves are said to have a Contact of the Third Order ; and 
so on. Hence, generally, if 

4> (*)=*(«), #'(*) =f (4 *"(*)=♦» • • • <r( a )=r (*) . • • (*) 

the curves have a contact 0/" //;<? n th order. 

Cor. i . A contact of the n th order involves n -f- i conditions. 

Cor. 2 . As only n + i conditions can *>z general be imposed 
upon a locus whose equation contains n -{- i arbitrary constants, 
the highest order of contact to such a curve is in general the 
n th . Thus the straight lines Ax -f- By + C = o, having only 
two arbitrary constants, can have a contact of the ist order ; the 
circle (x — a) 2 -f- (y — <£) 2 = ^ 2 having three arbitrary constants 
can have a contact of the second order. 

142. Two curves in contact do or do not cross each other at their 
co7?imon point according as their order of contact is even or odd. 




Let y = <f> (x) and y = \p (x) be the equations of two curves 
having the n th order of contact, and let x = a be the abscissa 
of their common point P, Fig. 30. Let us add a small incre- 
ment // to a, then 

<f> (a -\- h) — tfj (a + h) 

is the difference between the ordinates of the two curves corre- 
sponding to the same abscissa x = a + h. Expanding both 



194 Differential Calculus 

terms in the expression by Taylor's Theorem, and collecting 
like derivatives, we have (since <£ (a) = \j/ (a)) 

$(«+*)- f (a + h) = 1 4>'(a) - f(a) \h + \ 4>"(a) - ^\a) ] ~ 

\i 

+^'»-f»j^+ ( t) 

i . If n is even, 

then 4>'(a) = i//(V), <£"(") =* tf'(a), . . . cf> n (a) = f 1 (a), 

and the terms in the second number of (i) successively vanish 
until the (// + i) th term is reached. As this term contains h 
affected with an odd exponent (// + i) its sign will change with 
/i, and if h is taken sufficiently small the numerical value of this 
term will exceed the sum of all the other terms. Hence the 
sign of the second member, that is, the sign of the difference of 
the ordinates, changes with the sign of h ; hence the curves 
cross each other. In Fig. 30, the curves MN and S' T' illus- 
trate this case, P being the point of contact. 

2. If n is odd, 
then the first term that does not vanish contains h affected with 
an even exponent (;/ + 1). If this term is made to control in 
the second member by giving h a very small value, then the 
second member, and hence the difference of the ordinates, will 
not change sign with h ; hence the curves do not cross each 
other. 

The curves J/TVand ST illustrate this case. 

Cor. 1. Since the smaller the difference between the ordi- 
nates <£ (a + K) and \p (a -f- h) for any small value of h the closer 
the curves approach coincidence near P and since <£ (a + K) 
— \p (a + /j) becomes smaller and smaller as the number of 
terms in the second member of (1) decreases, it follows that 
as the order of contact increases, the c/oser the curves approach 
coincidence. 



Contact of Curves Envelopes 195 

143. At a point of maximum or minimum curvature the circle 
of 'curvature has a contact of the third order. 

Let y=f(x) (<i) 

and (x - a) 2 ■+■ (y - bf = r 2 (/>) 

be the equation of the curve and circle of curvature, respec- 
tively. 

From § 134 (1), we have 

d 2 y 

dx 2 



By condition, 
Hence, 



Mx 



dc 

~r = °« § 112. 

dx 



dc } \dxj ) dx z dx 2 2 ) \dx) ) dx dx 2 

dx c fdyY ) 3 

( \dx) C 

dy (d 2 yy 



Solving, we have — 3 = — - . 



Differentiating (//) successively, we have 

/ n dy 

x -a + (y-o)— = o, 

' ^/jc 3 dx dx 2 dx dx 2 



Substitutir 


lg the value of 


y — b drawr 


with respect 


d z y 
dx*' 


we have 








d\y 


dy td 2 y^ 
3 dx [dx 2 / 






dx z 


dy 2 
dx 2 



196 Differential Calculus 

b drawn from (</), and solving 



<<) 



which is identical with (c). The third derivatives are therefore 
equal, and the contact is of the third order. 

We may obtain the same result in the following very simple 
manner. Since at the point the curvature is a maximum or a 
minimum the circle of curvature does not cross the curve, §135 ; 
hence, § 142, 2, the order of contact is odd. But the expression 

fy 

dx 2 



c = 



is the expression for both the curvature of the circle and the 
curve at the point of contact; hence, thejirst and second de- 
rivatives drawn from their equations are equal ; hence, the order 
of contact of the curves is, at least, the third. This article and 
the one which follows explain the significance of the term " in 
general," used in § 141, Cor. 2. The general statement there 
given admits of exceptions at certain singular points of curves. 

144. The tangent to a curve at a point of inflexion crosses the 
curve. 

Let y =/(x) (a) 

and Ax + By -f- C = o (b) 

be the equations of the curve and its tangent at a point of 
inflexion. 

Then, we have from (a), 

d 2 y __ 

dx 2 



Contact of Curves Envelopes 197 

as a condition for a point of inflexion, § 128. From (b) we have 

also, 

d 2 y _ 
d*x~°' 

hence, at a point of inflexion, the order of contact of the curve 
and its tangent is the second ; hence the tangent and the curve 
cross each other, § 142, 1. See Fig. 23. 



EXAMPLES. 

1. Find the order of contact of the curves, 

(a) y = 2 x 2 , and (Ji) y = 3 x — x 2 . 

Here (1, 2) is their common point. Differentiating and sub- 
stituting, we have, 

from (a) — = ax = 4, from (fr) — - = 7. — 2 x =\. 

v J dx v ; dx ° 

.*. The curves intersect at the point (1, 2). 

2. Find the order of contact of 

y = ax 3 , y = 3 ax 2 — 3 ax -f- <z. 

Combining the equations we have (1, a) for the common 
point ; then 

-= 3 ax 2 = 3 a, 

d2 y c * 

— — - = o ax = o a, 
dxr 

Hence the curves have the second order of contact. 

a 2 

3. What order of contact has the circle,(# — § df + ( y — %df = — > 



—- = ax - 
dx 


- 3 a = 3 a - 


^=6 a. 

dx 2 




d 3 y _ 
dx 3 ~ 





and the parabola, x*+. y*= a&, at (- , — y 

\4 4/ 



^//x. Third. 



198 Differential Calculus 

4. Find the equation of the circle of curvature of the curve, 
j' 4 == 4<2 2 jc 2 — x*. 

5. What is the highest order of contact possible to two 
conies ? Ans. Fourth. 

6. What is the highest order of contact possible to the ellipse 
and parabola. Ajis. Third. 

7. Given xy=^x — i and jy— x— 1 = a(x — i) 2 , find the value 
of a in order that the two curves may have a contact of the 
second order. 

145. Families of Curves. Curves whose equations differ o?ily in 
the values of the constants which enter them are said to be of the 
same family. 

Thus the equation (x — df -\- (y—b) 2 = r 2 is the equation of a 
family of a circles whose positions and magnitudes depend upon 
the values of the constants a, b, r. Again, the equation Ax + 
By -\- C = o is the equation of a family of straight lines whose 
directions and positions with reference to the axes depend upon 
the values of A, B, C. 

The constants which enter equations are called Parameters, 
and if one or more of these are supposed to vary, they are 
called Variable Parameters. 

146. Envelope. The Envelope of a family of curves is a curve 
tangent to each ?7iember of the family. 

Thus, if we assume a to be the variable parameter in the 
equation (x — a) 2 + (y — b) 2 = r 2 , b and r remaining constant, 
we have (Fig.- 31) a* series of circles, all of whose centers are on 
the line MJV, at a distance b from the ^-axis. The envelopes 
of this family of circles are evidently the || lines AB and EF, 
whose equations are 

y = b ± r. 



Contact of Curves Envelopes 



199 




If we assume b to vary, a and r remaining constant, we have 
a family of circles whose centers lie along the line LK at a dis- 
tance a from the jy-axis. 
The envelopes in this 
case being CD and 
HG, whose equations 

are 

x = a ± r. 

Similarly, we 
suppose a and 
vary, r remaining con- 
stant, or a and r to 
vary, b remaining con- 
stant, etc. ; or we may 
suppose all three to 
vary at the same time. 
In each case we have 
a family of circles, and a curve tangent to the members of that 
family is called the envelope. 

It is evident, in this case, that if r alone varies there is no 
envelope. 

147. To determine the equation of the envelope. 

Let u =f(x, y, a) = o . . . (V) be the equation of any one 

(MJV) of a family of curves, 
a being the variable param- 
eter, and let u =f(x, y) = 
o . . . (7/) be the equation of 
the envelope ST. Let P 
(x, y) be the point of tan- 
gency. 

Since the curves are tan- 
Fig. 32. 1 ! / N 

gent to each other at (x, y) 
the first derivatives drawn from their equations must be equal ; 
hence differentiating each we have, 




200 Differential Calculus 



. du du du 

du = — dx -\ — r dy -\ — — da = o. 
dx dy da 

du du da 
dy dx da dx 



dx du 

dy 

a i r / 7N 7 du 7 du . 

Also, from (a), du = — ^/x -f — dy. 

dx dy 

du 
dy dx 

dx du 

~dy 



(") 



if) 



But (e) and (/*) are equal ; hence equating and reducing, we 

have 

du da 

da dx 

du 

da 



= ° (g) 



is a condition which the envelope must fulfill. If, then, we com- 
bine ( < § r ) and (V), eliminating a, we have an equation expressing 
the relation between x and y for every such point as P\ hence 
the equation, thus ascertained, is the equation of the envelope. 
To illustrate, let us find the envelope of the family (x — a/ 
+ (y — by = r 2, in which a is the variable parameter. 

Here u = (x — df -f- (y — frf — r 1 = o ; 

du 
.'. — = — 2 (x — a) = o ; 

.•. a = x. 

Substituting this value of a in (x — df -\- (y — b) 2 = r 2 , we 

have 

y = b ± r, 
as before. See § 146. 



Contact of Curves Envelopes 201 

Similarly we may show that 

x = a ± r 

is the equation of the envelope when b is the variable param- 
eter. 

148. The evolute of any curve is the envelope of its normals. 

We might readily infer this from § 140. We may prove the 
fact, however, as follows : 
From § 70, we have 

y-y = -j/( x ~ x )> 

, ,. dy r . , v 

or, (y — y ) —-, + x — x = o (a) 

for the equation of the normal to any plane curve whose equa- 
tion is expressed in Cartesian coordinates. Taking x as the 
variable parameter and differentiating, we have 

, <*y d/* 

Solving (p) with respect to y and substituting in (a), we have, 



" y*7 



2,/ 



dx 7 * 

f dy r V\ dy' 



r I \dx' J ) dx r 

x = x ly 7 

dx f2 

for the coordinates of any point on the envelope. But these 
values are identical with the coordinates of the center of curva- 
ture as found in § 138. Hence the envelope of the normals is 
the evolute of the curve. 



202 Differential Calculus 



EXAMPLES. 

P 
1. Find the equation of the envelope of the lines, y = sx + — 

2 j- 



i" being the variable parameter. 



Here u =y — sx — 



2 J - 



1 ^7/ ^ c P 

hence, — - = — x + ^ = o : .*. .r = -=— ; 

7?.f 2r 2 .* 

i.e., y 2 = 2/jf, 

a parabola. The given equation y = sx -\ will be recog- 
nized as the slope form of the equation of the tangent to the 
parabola ^ = 2px. See Ana. Geom., p. 96. The parabola is, 
of course, the envelope of its tangents. 

2. Find the envelope of the family of lines represented by 
each of the following equations : 



y = SX ± A \/s 2 a 2 + ^, y = SX ± \ls 2 (l 2 — b 2 . 

x 2 y 2 x 2 y 2 

-2 + j2 =1 ' a 2 ~l? 



Aus. — -f — _i, — — — — 1. 



3. The slope form of the equation of the normal to the parab- 
ola y 2 = 4 ax is y = (x — 2 a) s — as? ; find the equation of 
the evolute of the curve. By § 148, the evolute is the envelope 
of the normal. We are required, therefore, to find the envelope 
of the series of lines represented by the equation y = sx — 2 as 
— as 3 , s being the variable parameter. 

du 
Here — = — x-\-2a-\-^as 2 = o- 1 



-4 



x — 2 a m 



3 a 



Contact of Curves Envelopes 



203 



y = (x — 2 a) y - 

i — 
= 3"\ 



— 2 a 



3 a 



■v 



x — 2 a 



3" 



"-)•-< 



x — 2 a 



3 a 



— vp^y 

.-. 27 ay 2 = 4 a 2 (x — 2 a) 3 , 

which is the desired equation. We have previously deduced 
this equation by the direct method. See § 138 and figure 28. 

4. The hypotenuse of a right 
triangle changes its position, 
its length remaining unaltered ; 
find its envelope. 

Let OB A be the triangle, 
BA being any one position of 
the hypotenuse. 

Let BA = c, a constant, OB 
= b, OA — a. Then the equa- 
tion of BA is 




Fig. 33. 



and by condition, 



x y 
a 
a *+P=c 2 






Let us take a as the variable parameter. Ordinarily we 
would find the value of b in terms of a from the given condi- 
tion, and substitute in the equation of the line, and then pro- 
ceed as in preceding examples ; but in this case, as in others 
with which we have had to deal, the simpler process is to sub- 
stitute after differentiation. 

Since b is a function of a, we have from (;#), 

du x y db 



da 



a 



b 2 da 



= o 



(/) 



204 Differential Calculus 

from (///), 2a -{- 2 o— = 0; 

da b 

This value in (/) gives, after reduction, 

Jpx = a z y (r) 

We are now to eliminate a and b, having the relation (/«), 
(/z), and (r). 

From (//) and (r) we find, 

ex* cx^ 

a = - T £ = 



These values in (;;/) give, after reduction, 

X5 -\- y* = c-i 

for the equation of the envelope. This curve is the four-cusped 
hypocycloid, and is generated by a point on the circumference 
of a circle as it rolls on the concave side of another circle whose 
diameter is four times that of the rolling circle. This problem 
was discussed by John Bernouilli in 1692. 

x y 

5. Find the envelope of the lines — f- — = 1, subject to the 

a b 

condition that 's/a + V^ = V^, a constant. 

Ans. x^ -\- y* = c*. 

6. Prove by direct process that the envelope of the lines 

xy 

— \- ^ = 1, subject to the condition ab = 2 c is 2 xy == c, a hyper- 

a b 

bola. 

7. Find the envelope of all ellipses having a constant area 
(7r£ 2 ), the axes being coincident. Ans. 2 xy = ± a. 



Singular Points 205 



CHAPTER XV. 

SINGULAR POINTS. 

History. — Joseph Saurin (1659— 1737) was the first to show now the 
tangents at multiple points of curves could be determined by analysis. 

Newton discussed double points in a plane and at infinity in his " Optics ' 
(1704). 

Rules for finding and discriminating multiple points were given by Mac- 
laurin in his "Treatise of Fluxions" (1742). 

149. Singular Points are those points of a curve having some 
peculiar property not possessed by the other points of the curve. 
Thus, the point of inflexion is singular in that it is the point 
where the direction of curvature changes, — a property not pos- 
sessed by the other points of the curve. With this point we 
have already had to deal (§ 128). It is now our purpose to 
consider in order the more common of these singular points. 

150. Multiple Points are those points of a curve common to two 
or more of its branches. 

As the branches must either pass through the point or simply 
meet at the point, there are two classes : 
1. The branches pass through the point. 





(2) 

Fig. 34- 

(a) If the branches inte7'sect, the point is called a Point of In- 
tersection ; and the point is a double, triple, or quadruple . . . point 



206 



Differential Calculus 



according as two, three, or four . . . branches pass through it. 
Figures i and 2 illustrate a double and a triple point. Since the 

curve has as many tangents at a point of intersection as there 

dy 

are branches, it is obvious that — must have two different val- 
ues at a double point ; three different values at a triple point ; 
and so on. 

(b) If the branches touch as they pass through the point, it is 
called a Point of Osculation ; and this point is of the first species 
or second species, according as the branches lie on opposite or on 
the same side of their common tangent. 




Fig. 34- 

Thus, Figs. (3) and (4) are illustrations of osculating points 

dy 
of the first and second species. Here — has two equal values. 

2 . The branches 7neet at the point. 

(a) If the branches have a common tangent at the point, the 
point is a Cusp of the first or second order, according as the 
branches lie on opposite or on the same side of the tangent. 





Fig. 34. 



(6) 



dy 



Here, also, — has two equal values. 
dx 




Fig. 34 (7)- 



Singular Points 207 

(f) If the branches have different tangents at the point the 

point is called a Point Saillant, or Shooting-Point. 

dy 
Here — has two differe7it values, It 

may be remarked, however, that shoot- 
ing-points occur only in loci whose equa- 
tions are transcendental. Such points 
are usually determined by inspection. 

151. Isolated or Conjugate Points are 

those points which are isolated from the 

curve, but whose coordinates satisfy its equation. 

As the curve has no direction at an isolated point, -P(x, y~), 

dy 
Fig. 8, it is obvious that — has an imaginary 

value at such a point. But imaginary values 
arise from the presence of radicals with even in- 
dices ; hence, if — has one imaginary value it p. 
has necessarily two such values. 

152. The Point d' Arret, or Stop Point, is a Fig. 34 (8). 
point at which a branch of a curve stops. This 

point, peculiar to transcendental curves, is usually determined 
by inspection. 

153. Investigation for Singular Points. Let u =f(x, y) = o 
be the rationalized equation of any plane locus ; then §121, 
Cor. 1, 

du 
dy dx 

dx du 

dy 

dy 
We have found (§§150, 151) that —must have more than one 

value in all cases of multiple and isolated points. But, since 




208 Differential Calculus 



differentiation of a rational equation cannot give rise to an irra- 
tional expressi 
(V, _/) unless 



tional expression, — can have only one value for any given point 



dy o m 
dx o 

, du du 

i.e.. unless -=- = o, and — - = o. 

dx dy 

Hence, since the point (x',y') satisfies the equation,/"^, y) = o, 

du du 

if u = o, — = o, — = o, 

•or dy 

for that point then it maybe a multiple or isolated point; i.e., x' 
and y r are critical values which require investigation. Further 
investigation consists in evaluating the expression 



du- 
dy~\ dx 

dx\ (x)yf) du 

dy 



o 
o 

(a/, ?/) 



Referring now to the figures and definitions of §§ 150, 151, 
we see : 

dy' 

(a) If — — has two or more unequal values, and y is real for 

x = x' ± /?, h being a small increment, the point is a point of 
intersection. See § 150, 1, (a). 

dy' 

(b) If -=— t has two or more unequal values, and y is real for 

x = x r -\- h (or x = x r — n), and imaginary for x = x — h (or 
x = x f -\- /i), the point is a shooting-point. § 150, 2, (Z>). 

dy' 
(e) If -^— , has two or more equal values, and y is real for 

x = x' ± h, the point is an osculating-point. § 150, 1, (£). 



Singular Points 209 

dV 

(d) If —. has two or more equal values, and y is real for 

v 7 dx 

x = x' -f- h (or x = x' — h) and imaginary for x = x' — h (or x 
= x' -\-h), the point is a cusp. § 150, 2, (a). 

To determine in the last two cases, (c) and (d), whether the 
point is of the first or second species (Figures (3), (4), (5), (6)), 
usually the simplest way is to write the equation of the tangent 

dV 
y — y' = -j-, ( x ~ x )> an( i compare the ordinates of the tangent 

for x = (x f + X) or x = x' — h with those of the curve for the 
same abscissa. If the ordinate of the tangent exceeds (or is 
less than) the corresponding ordinates of the curve, the point is 
of the second species ; otherwise it is of the first species. 

(e) If -j- f has an imaginary value, and y' is real when x = x', 

the point is isolated. 

While the above statement is true, yet the converse, viz., that 

dyf . 
at an isolated point (x r , y r ), ——, is imaginary, is not necessarily 

true. For at such a point y is necessarily imaginary, when 
x = x' ± h (/i being some small quantity). Now, by Taylor's 
Theorem, we have 

y ='/<V ± K) =/<V) ±/'(V) h +/"(*') I ± ■ • • 

Hence, y is imaginary when any derivative in the second 

member is imaginary. 

dy 
Hence, y may be imaginary while — ( y =f'(x'' s j) may be real. 

In examining, therefore, a given curve for isolated points the 
simplest and most satisfactory test is, after determining the 
critical values as above explained, to substitute values (x ± n) 
a little less and a little greater than these in the original equa- 
tion, and ascertain if they render y imaginary. 



210 



Differential Calculus 



EXAMPLES. 
1. Investigate the lemniscata for singular points. 
Here, 



and 



u = (x 2 + ff - (fix 2 ~f) = o, 

— = 4 x (x 2 -f- jf) — 2 a 2 x = o, 

— = 4 y (x 2 -f- y 2 ) + 2 tf 2 _y = o. 
Solving the last two equations, we find, 

(o, o), FV2", oj, f-^VI, oj, 




Fig. 35- 



to be critical points. Of these, 
however, the first (o, o) only 
satisfies the condition u = o 

(§ 153). We are, therefore, to 

dy 
evaluate the first derivative — 

ax 

which takes the illusory form - 

for the point (o, o). 
That is, 



dy 
dx 



J 0,0 



dx 

bu 
dy_ 



4 x s -J- 4 x} 2 — 2 a 2 x 
4 x 2 y + 4-jfi + 2 a 2 y 



0,0 



hence, 



0,0 



dy 

dx 



12 x 2 -f- 4(2 ocy— — \- f-\— 2 a 2 

a[x z — — (- 2 yx)-\-i2 y— — h 2 or — - 
\ dx J dx dx 

.'. ( — J = 1 when x = o and 7 = 0; 



2 a' 



2 a c 



0,0 



dx 



dx 



= ± 1, 



Singular Points 



211 



Therefore the curve has two non-coincident tangents at the 
origin. Therefore, since the equation of the curve is algebraic, 
the origin is a double point of intersection. The values of the 
derivative (± i) show that the tangents are inclined at angles 
of 45 and 135 to the *-axis. 

2. Show that the curve ^ = x* (1 — x 2 ) has a double point of 
osculation of the first species at the origin. 

Here, 



u = 


f- 


X 4 + X 6 


=: 


O, 


du 
dx 


6x* 


-4^ = 


O, 




bu 










dy 


2y- 


= 0. 








Fig. 36. 

We see that the partial derivatives give (o, o), a point of the 
curve, as a critical point ; 

dy 6 x 5 — 4 x z ~ 

dx 



hence, 



dy 
dx 



2 y 
30 x 4 — \'zx L 



JO.O 



dx 



0,0 



o 

~dy 

dx 



IT- 



dx 



= ± o for the point (o, o). 



Hence, the curve has two coincident tangents at the origin, 
which coincide with the ^-axis. Hence the point (o, o) is a 
double point of osculation, or a cusp. 

Resuming the equation, f- = x 4 — x G , we have, 



y = ^ a 2 vi 



x 1 



which, for all values of x less, numerically, than 1, give real 
values for y; hence § 153, (V), (o, o) is a point of osculation. 

Again, since the equation shows that the curve is symmetrical, 
with respect to the A>axis, i.e., to its tangent at the origin, the 
origin is a point of osculation of the first species. 



212 



Differential Calculus 



3. Show that the curve y 2 — x z {\ — x) has a cusp of the first 
species at the origin. 



u = y* — x z -f- oc 



4 




du 

dx 
du 
dy 



= — 3 x? + 4 x s 



= 2 y = o. 



Fig. 37- 



.*. (o, o) is a critical point. 





du 








dy 


dx 


3 - 5>n 

4 x a — 3 ^r 


o 


dx 


du 


2 J' 


0,0 ° 




dy 


0,0 


dy 


i 


2 JT 2 — 6 X~ 


O 


dx 




2 ^ 


dy 

2 






*/.# 


0,0 dx 


\dx) 


o. 


^ 
rt^ 


: o, at the origin. 



Therefore, at the origin (o, o), the curve has two tangents 
coinciding with the .r-axis. 

From the equation of the curve, we have, 

y = ± V^ 3 (i — x). 

Hence, since x cannot be negative, the curve is situated in 
the first and fourth quadrants, and is symmetrical with respect 
to the x-axis. Hence the origin is a cusp of the first species. 

4. Show that the cissoid (2 a — x) f = x 5 has a cusp of the 
first species at the origin. 

5. Show that (y — x 2 ) 2 = x* has a cusp of second species at 
the origin. 

6. Show that the semi-cubic parabola ay* = x z has a cusp of 
the first species at the origin. 



Singular Points 



213 



y i 

7. Show that the cycloid x = a vers -1 V2ay—y 2 has an 

J a 

infinite number of cusps of the first species. 

8. Show that the curve a (x 2 + f) = x s has a conjugate 
point at the origin. 

u = ax 2 -f- ay 2 — x 3 = o. 
du 



dx 



= 2 ax — 3 x 2 = o. 



.*. (o, o) is a critical point. 




dy 

dx 



3 x z — 2 ax 
2 ay 



6 x — 2a' 



0,0 ^ 



Fig. 38 



I 

dy 
dx 



JY= - I; . or £ = ± V=I. 

Hence, the origin is a conjugate point. See (<?) § 153. 
Otherwise, thus : Solving the equation, we have, 



y z= -j- x 



x — a 



a 



This equation is satisfied for the point (o, o). But y is 
imaginary for any negative value of x and for any positive 
value of x less than a ; hence (o, o) is isolated from the curve. 

9. The curve y 2 = x (x + df has a conjugate point at 
(— a, o). 

10. The origin is a conjugate point of the curve y 2 (x 2 — a 2 ) 

4. 

= x . 

11. Show that the point (a, o) is a conjugate point of the 
curve ay 2 — x z -f- 4 ax 2 — 5 <z 2 .# + 2 # 3 = o. 



214 



Differential Calculus 



12. Show that the curve a 3 y 2 — 2 abxy — x r ° = o has a double 
point of osculation at the origin and that one branch of the 
curve has a point of inflexion at that point. 

13. Show that the curve y = ^cot -1 ^ has a point saillant 

at the origin. 

Since y is positive, and has only one 
value for all values of x, positive or 
negative, the curve lies in the first and 
second angles ; and since x = o gives 
y = o, the curve passes through the 
origin. 




Fig- 39. 



Here 



dy 
dx 



= cot 1 X 



X 



1 + X 2 



If we suppose x to approach o from the positive direction, 
we have, 

7T 



dy 
dx 



cot 



1 x = cot * o = - = 1 . q 7 . 

If we suppose x negative and approaching o, we have 

= cor 1 (- o) = - ^ = - 1.57. 



_£ =cot - 1( _ x)+ __ 



Hence there are two non-coincident tangents to the curve at 

the origin. Hence the origin is a point saillant. § 150. 

1 

14. Show that the curve jy — x + ye x = o has a point saillant 

at the origin. 

1 

15. Show that the curve y — e x = o has a point d 'arret at 

the origin. 

16. Show that the curve y = x \ogx has a point d'arret at 
the origin. 

17. x* — ax^y — axy* + a 2 y 2 = o. 

A conjugate point at (o, o). 



Singular Points 215 

18. x 4 -f- 2 axp-y — ay 3 = o. 

A triple point of intersection at (o, o). 

19. af = (x — df (x — b). 

At x = a there is a conjugate point, a 
double point or a cusp according 
as a < b, a > b or # = b. 

20. Examine the hypocycloid x z -\- y* = a 3 for cusps. 



216 Differential Calculus 



CHAPTER XVI. 

LOCI. 

154. In tracing curves in Analytic Geometry we usually solve 
the equation of the curve with respect to one of the variables 
that enter it ; then assigning values to the variable in the 
second member we determine the values of the other. A 
smooth curve traced through the points thus determined we 
call the locus of the equation. This process is at best only 
approximate and is limited in its application to those curves 
whose equations are of lower degrees. In equations of higher 
degrees the difficulty is even greater as we can only determine 
approximately the positions of the series of points. By the aid 
of the Differential Calculus we are enabled to determine the 
singularities of the locus from its equation and from these to 
obtain a general idea of its form. We have seen in the pre- 
ceding chapter, for example, how to investigate any locus for 
singular points; in § 127, how to determine the direction of 
curvature; in §§ 73, 79, how to determine whether or not the 
curve has asymptotes and if so, to determine their equations, 
etc. We propose to treat a few curves in this general manner, 
and to indicate an order of procedure that will enable the stu- 
dent to enter upon an intelligent investigation of any equation 
with which he may have to deal. 

ALGEBRAIC EQUATIONS. 

155. Suggestions. 

1. Determine as far as possible the form and properties of 
the locus from its equation. 



Loci 217 

2 Deduce the first and second derivatives from the equation 
and investigate. 

(a), for asymptotes. Cf. §73. 

(J?), for maxima and minima points. Cf. § 114. 

(<r), for singular points. Cf. § 153. 

(d), for direction of curvature. Cf. § 127. 

EXAMPLES. 

a 2 x 
1. Trace the curve y = 



(x — df 

Here x = o gives y = o ; hence the curve passes through 
the origin. As x approaches the value a, y approaches an in- 
finite value ; hence x = a is the equation of an asymptote to the 
curve, § 74. Again as x increases numerically, and approaches 
positive or negative infinity, y decreases and approaches o as 
a limit ; hence y = o, or the j*>axis is an asymptote to the two 
branches of the curve, one extending infinitely in the first angle 
and the other infinitely in the third angle. From the given 
equation we have 

dy a 2 (x + a) , . d 2 y 2 a 2 (x -f- 2 d) /7 . 

dx (x — df v ' dx 2 (x — ay 

dy . . . 

Here — — o gives x = — a, and this value in (Ji) gives 

d 2 y 1 . . . 

— 2 = - — , a positive quantity ; hence at the point whose ab- 

scissa is — a, y is a minimum. 

dy 
Making x — o in (a) we find — = 1 ; hence the tangent at 

the origin (since x — o gives y = o) makes an angle of 45 ° with 
the ^r-axis. 

d 2 y 
Placing —— = o, we find x = — 2 a. Since (b) changes sign 

as x passes through this value, x — — 2 a is the abscissa of a 



218 



Differential Calculus 




Fig. 40. 



d 2 y 
point of inflexion. Again, since -^ is positive or negative 

according as x is algebraically greater or less than — 2 a, the 

curve is concave upward be- 
tween the limits x = — 2 a 
and x = 00 , and concave 
downward between the limits 
x = — 2 a and x = — 00 . 
Reviewing the facts elicited 
we are enabled to trace the 
curve as in the figure. It 
may be remarked in passing 
that an asymptote to an al- 
gebraic curve is always approached by two infinite branches. 
The hyperbola affords a familiar illustration. 

2. Trace the curve y 2 = x 2 (x — a). 

Here x = o gives y = o ; hence the curve passes through 
the origin. 

Again, all negative values of x, and all positive values of x 
less than a, render y imaginary ; hence the origin is an isolated 
point. 

All positive values of x greater than a give two values of y 
equal numerically with contrary signs ; hence the curve is sym- 
metrical with respect to the .x-axis and extends indefinitely in 
the direction of positive abscissas from the limit x = a. When 
x = a,y = o ; hence the curve cuts the ^-axis at the point (a, o). 

Here 
d y 3 X — 2a , x d % y 3 X ~ \ a 

—a ' ; • ' W; ~d^~ 
dy 



dx 



\[x 



(*) 



When x = a, 



dx 



4 V(^ — of 
= 00 ; hence the tangent to the curve where 



it crosses the jc-axis is perpendicular to that axis. 

Since (a) does not change sign as x increases, the curve has 
no maximum or minimum points. 



Loci 



219 



Since (J?) changes sign as x passes through the value f a 
there are points of inflexion corresponding to this value of x. 

If we take the positive sign of the radical in the denominator 

d 2 y 
of (p) we find that — ^ is negative 

between the limits x = a and x = 
| a, and positive for all values of 
x greater than ^ a ; if we take the 
negative value of the radical we 
find the reverse is true. Hence the 
curve is concave toward the x-axis 
between the limits x = a and x = 
§ a, and convex toward that axis when x > f a. 

3. Trace the curve a z y 2 = 2 a^x^y + x 5 . 




Fig. 41. 



Solving we have y = —^ (a ± Vtf (x + a)). 



x = o gives y = o ; hence the origin is a point of the curve. 

For all positive values of x, y has two real values of opposite 
signs ; hence the curve extends indefinitely in the first and fourth 
angle, x = — a, y = a, and for all negative values of x be- 
tween the limits x = o and x = — a, y has two positive unequal 
values ; for negative values of x greater numerically than — a 
y is imaginary ; hence the curve has a loop in the second angle. 

Let u = «y — 2a 2 x 2 y — x 5 = o ; then 



Evaluating, 



hence, 



dy 
dx 



dy 

dx 

1 

dy 

dx 



4 a 2 xy + c x* o . . . . 

— ~ V-9 = - for tne point (o, o). 

2 cry — 2 a z x* o v ' 



4 arx — — \- 4 <3rv -f 2or 

3^ 2 

2 ar — 4 # ^ 

#x 



2 #' 



_o, 



dy" 
dx 



= ±0; 



.-. The origin is a point of osculation, the x-axis being a 






220 



Differential Calculus 



common tangent to the two 
branches. But we have seen 
above that one branch of the 
curve crosses the :v-axis at the 
origin (i.e., the curve has been 
shown to pass from the second to 
Fig. 4 2. the fourth angle through the 

origin) ; hence the origin is also 

a point of inflexion. Such a point is called a point of oscul- 

inflexion. 

4. y 





Fig. 43- 



5. y -f- xy = x. 




Fig. 44. 



6. x s — 2 x 2 y — 2 x 2 — Sy = o. 




Fig- 45- 



Loci 221 

7. y (x — a) = x (x — 2 a). 

8. ay 2 = x* + x 5 . 

9. y (a — x) = x 2 (a + x). 

10. y = x 2 (i - x 2 ) s . 

11. /(^- tf 2 ) =x\ 

12. j^ 3 = ^ 3 — x 3 . 

13. ^(^ 2 -^) = « 3 . 

POLAR EQUATIONS. 
156. Suggestions. 

1. Determine as far as possible the form and properties of 
the curve from its equation. 

2. Deduce the first derivative of r with respect to from 
the equation of the curve. 

(a) Investigate for asymptotes, Cf. § 79. 

(If) Investigate for maximum and minimum points. Cf. 

§ 112. 
(<r) Investigate for points of inflexion. Cf. § 131. 
(d) Investigate for direction of curvature. Cf. § 130. 

EXAMPLES. 
1. Trace the curve r = a sin 3 6. 

r = o, when = o°, = 6o°,0 =120°, = 180 , etc. ; hence, 
the curve repeatedly passes through the origin. 

r = a (a maximum value, since sin 3 cannot exceed unity) 
when 6 = 30 , = i 5 o° ; 6 = 270 . 

r = — a, a minimum value, when = — 30 , = — 150°, 
= - 270 . 

As increases from o° to 30 , r increases from o to a ; as ^ 
increases from 30 to 6o°, r decreases from a to o ; hence, the 
curve has a loop in the first angle. 

As 6 increases from 6o° to 90°, r decreases from o to — a ; as 
increases from 90 to 120 , r increases from — a to o ; hence, 



222 



Differential Calculus 



the curve has a similar loop to the first, situated partially in the 
third angle and partially in the fourth angle. 

As increases from 120 to 150 , r increases from o to a ; as 
increases from 150 to 180 , r diminishes from <? to o ; hence, 
there is a loop in the second angle. 

As 6 increases from 180 to 360 , the corresponding values of 

r are the same in magnitude and direction as those already 

indicated. 

-r-r d r n 

Here, — = 3 a cos 3 0. 



dO 



dr 



Since ^ = 3 a cos 3^ = when = 30* 
dr 



*5° 



= 270 



and since — = changes sign as passes through these values it 

follows that r is a maximum for these values of 0, — a fact 
already ascertained from the equation. 




Fig. 46. 



2. r = a sin 2 0. 




Fig. 47. 



Loci 



223 



3. r 2 = tf 3 cos 4 f 6. 




4. r 4 = a 5 cos 5 % 6. 







Fig. 4 8. 




Fig. 49. 



5. r = a sec-- 
3 
From the equation we readily see that the curve is of the gen- 
eral form given in the figure. 



Here 




Fig. 50. 

. 

dr a 6 a 2 
— = - sec - tan - = * 

dv 3 3 3 3 o# 
° ° J cos 2 - 

3 
Subtangent — r* — = 3 # esc - • 



</r 



224 Differential Calculus 

When = 270 , we have, 



and 



r = 00 , 
Subtangent = 3 # ; 



hence, a line perpendicular to the initial line, and at a distance 
3 a to the left of the pole, is an asymptote to the curve. 



6. r = 2 a tan sin 0. (Cissoid). 

sin 2 2 a 



Hence, 



r = 2 a 



cos cos 



2 a cos 0. 



= o°, r = o ; and as increases, r increases. 
When 6 = 90 , r = 00. 

As decreases from o° to — 90 , r increases 
from o to 00. 

From the equation above, we have 

dr 2 a sin 6(2 cos 2 6 + sin 2 0) 
d0 = cos 2 (9 

o //0 2 tf sin 3 
Subtangent = r 

dO 




dr 2 — sin 2 
2 <? ; hence, a line J_ to the 



initial line, and at a distance 2 a X.o the right of the pole, is an 
asymptote. Tracing the curve from the above data, we find it 
as in the figure. 

7. r = a sec ± a. (Conchoid.) 

. 

8. r = a sin — 

2 

9. r = a cos 2 0. 
10. r = cot cos 0. 



PART II. 



INTEGRAL CALCULUS. 



INTEGRAL CALCULUS. 



CHAPTER I. 

TYPE FORMS. 

History. — The Integral Calculus may be said to have taken its origin 
from methods employed by Cavalieri, Wallis, and others for the determina- 
tion of quadrature of curves and cubature of solids. The processes thus 
employed were developed and reduced to a suitable notation by Newton and 
Leibnitz. 

The term " integral" was first used by James Bernouilli (1654-1705). 

157. The Integral Calculus is the inverse of the Differential ; 
and its fundamental object is to determine the function — the 
relation between the rates or differentials of the variables 
which enter it being given. 

158. Integral. Integration. A function is termed the integral 
of its differential, and the process by means of which it is derived 
is termed integration. 

The process of integration is simply a reversion of the process 
of differentiation ; hence no new philosophical principle is in- 
volved in the process. Thus, since 

3 o?dx is the differential of x 3 , 

x 3 is the integral of 3 x*dx. 

dx 
Again : — is the differential of log x ; 

dx 
hence, log x is the integral of — • 

227 



228 Integral Calculus 

159. Notation. The operation of integration is denoted by 
the symbol / , read " integral of." 

Thus, in the examples of § 158, we write 



S> 



I 



= log X. 



3 x 2 dx = x 3 , 

^dx 
x 

Since integration and differentiation involve inverse operation 

the symbols I and d neutralize each other. 

160. Indefinite Integrals. Constant of Integration. 

Since d (x 3 -f- 5) = 3 x 2 dx, 

d (x 3 — 3) = 3 x 2 dx, 
d (x 3 ± c) = 3 x 2 dx, 
and, in general, d {/(x) + c) =/' (x) dx, 
it follows that, 

I 3 x 2 dx = x 3 + c, 

\f'(x) dx —f(x) + c, 

where c is some indefinite constant. The constant c is called the 
Constant of Integration ; and as its value is in general unknown, 
the integral of which it forms a part is indefinite. While the 
process of integration gives, it seems, an indefinite result, yet in 
the practical application of the process the data of the problem 
will enable us to determine the value of c, or to eliminate it alto- 
gether, and thus enable us to render the result definite. 

To avoid useless repetition we shall omit the constant of in- 
tegration in what follows. The student must bear in mind, 
however, that it is to be understood as entering every integral 
expression. 



Type Forms 229 

161. Elementary principles. 

i . The integral of the sum of any number of differentials is the 
sum of their integrals . Cf. § 24, (1). 

Since d (u ± v) = du + dv :. I d (u + v) = j (du ± dv) ; 
i.e., I (du + dv) = u ± v. 

But I du ± I dv = u ± v ; 

.'. f(du ±dv) = j du± I dv («) 

The symbol of integration is, therefore, distributive. 
2. y4 constant factor may be placed before or after the integral 
sign. Cf. § 25, Cor. i. 

Since d (cu) = cdu .\ 1 d(cu) = I cdu\ 

i.e., I cdu = cu. 

But c J du = cu\ 

.'• I cc/tf = c J du (b) 

162. Type Formulae. 

1 . The integral of a variable with a constant exponent i?ito the 
differential of the variable is the variable with an exponent in- 
creased by one divided by the increased exponent. Cf. § 27, (9). 

Since d(u n + 1 ) = (n + 1) u n du .-. / d(u n + 1 ) = (n -f 1) / u n du ; 
i.e., (* +I )JW* = »-; 

i fdu= ii — (1) 

n -1- 1 



230 Integral Calculus 

2. The integral of a fractional expression in which the numera- 
tor is the differential of the denominator is the logarithm of the 
denominator. Cf. § 33, (14). 

Since d (log ti) = — 

Schol. Since 

/0 



/?-/ 



# x du ■=■ — — 00, 



we see that formula (1) does not apply when n = — 1. Such 
expressions, therefore, as I u~ x du should be placed in a frac- 
tional form I — and formula (2) applied. 

3. The integral of a co?istant with a variable exponent into the 
differential of the variable is the constant affected with the same 
exponent divided by the logarithm of the constant. Cf. § 33, (!3)- 

Since d (a") = a u \ogadu .'. J d (a u ) = log a j a u du ; 

i.e, k s .f«:,*-.r; 



I 



a 



log a 

Cor. If a = e, we have from (3), 



a" da = ,— (3) 



/• 



e u du = e u (4) 

This also follows directly from the fact that d (e u ) = e u du. 



Type Forms 231 

EXAMPLES. 

The numbers and letters which follow the examples refer to 
the formulae of §§ 161, 162. 

1. I x s dx = — • (1) 

/dx 1 1 x 2 1 

—— = J x~* dx = — = 2 xr. (1) 

/dx . 

-=\ogx. (2) 

4. C(a+'bxfbdx = ( a + bx y - (1) 

5. f (« + £r) 2 <& = i C(a + &r) 2 .y* = (" + $*? . (^ ( r ) 

(x 2 + i) 2 */*r = v — — L. (£) } (j) 

8. JW* = i ja** 3 dx=-^-- (b), (3) 

9. \el x dx=l&\ 00,(4) 
10. C(a+Bfdx= ( a 1 rb T . ■ (i) 

+ ^ / # -3 dx — 5 I x~*dx = alogx 5 -f- 



. -v-3 



232 Integral Calculus 

12. / (i + x 2 ) (i 4- x) xdx = J \x + x 2 + x 3 + * 4 J 



rtk 



= -+ -+- + 
2 3 4 5 



13 



i- 



6^4-i2 a* — 8 7 /" V 12 8, 

dx = / < .r — 6 H ; w 



= 6 a: +12 log ^H 

2 X 



f 



X X 



14. / ' = / (m 4- x 4 ) *4.x 3 dx = 2 \lm -\-x*. 
J Vw 4 x 4 *J 

15. I <r (4 mx 4- 3 <? 2 ^ 3 )* (4^4-9 (fix 2 ) dx = f ^(4 wje + 3 <? 2 .x 3 )3. 

/' a 2x 

16. J (^ 2x 4- d x ) dx = — h 2 <?**. 

J v y 2 logtf 

• I- 



X JXX X 

(e ax 4- <P) dx = h ae«. 



Y dx = 



1 + \oga 
dx 

r dx r x 

, x log # */ log x 
I }cgx dx 



Jp 



j °— =/ lo s 



^x log 2 X 

x— ==— 5 

^ 2 



(2) 
(1) 



21 



) ! 



J- — 7 ^C = — — r ( (-2^—1 



2 A? 2 * 



^ (logJc) m+1 
22. / log™.* — = v s ; 

X #Z + I 



/r l>xdx I 

3 ™^ ?= - A log( 3OT -6^) = log^==== 



^) s 



Type Forms 233 

24. / -j a^xdx = § = • 

J ° 2 log a 

25. / me?'e r dx = m^ x . 

26. f 2 - ^ 2 ) - **** = - 1 / 0* - ^) f ( - 2 */&) = - ^~ X ^ , 

/ "3 .X 2 -I - 2 Jt -4- I 

27. I -£ - dx = log (x s + -* 1 4- x + i). 

J x 3 -h x 2 4 * + i sv^ ^y 

/(x m — a m ydx x m , , 
^ — = — (> m - 4fl m + # 2w logo:, 
x 2 ;;/ 

30. Ji^ = J log (* 2 4 |). 

31. I 2ax(- 2 -hi Jdx = — (p 2 + x 2 )l 

' rt&f = h x 4 2 log (# — i). 

JC — I 2 

J (a 4 bx n ) 



28 



29 



(<z 4 &* n ) m bn(i — m) 



r dx i fi^ ; , ± 

34. / = / x m dx s= (mx) m . 

I m—l m—1 J v / 

(x m 4 mxf J K } v y 

2 Vx m 4- mx 



m 



234 Integral Calculus 

/ x m-ij tl j (x m + mx) 
dx = v - • 
x m -f- mx m 

=== — dx = jc + V#M~4. 

V* 2 + 4 

, /•(V* 2 +4 + *) 2 v „ _ (* 4- V^+4) 



36. 



37. 



d& = 



39 



V* 2 + 4 

rt&p = / < i > dx = I dx — 2 J ■ 

2^+3 J ( 2^+3 ) J J 



2dx 



2X+3 



= x — log (2 j; 4- 3) 2 - 
2<Zdfc 



/2 # #.* r 
1 - = J (2 ax — jv 2 ) - ^ 2 rtrjc -1 ^: 
3: \j2ax — x z J 



41 



= \ (2 ax x — 1) ^2 ax 2 dx = 

/xax 1 . . _ 3 i 

t — - = I (2 a — x) ?x zdx 
(2 ax — .ar)* J 

= I (2 tf.x — x — i) _ i^~Vx = 



2 V2 tf.r — ^r 2 



# 



VAr 2 4-3(V^+3 —x) 



a V 2 <^x — x 2 
= x -\- V^ 2 4- 3. 



43 



44 



45 



/ 



x n ~ x 4- -a? - *);/.- 



(#» 4- ^ w - 1 )^ 
(log .r) w ^c /« (log #) w + * 



q—p 

q(x n -\-x n ~ v ) 1 

* is ~ />) 



# 



^4-1 



jc — 2 



/ 4 

_ dx = 2 V* 4- — p* 



Type Forms 235 

46. I x y/x + adx = | - (x 4- #)* — %a(x -\- a)%. 

/dx 2 

-s]x + a + \[x 3<* 

/a 4- bx C\b ab' — b a ' ) 7 
— — dx = I < —. 4- _, . . -j—r l ax 
a' + b'x J (b r b'{a 4- b' x) ) 

b ab r — bd . . , ., . 
= J fX H ?* — g ^ + *)' 

163. Type Formulae. (Continued.) Cf. § 43. 

/sin udu = — cos a or vers a (5) 

/ cos u du = sin a or — covers a (6) 

/ sec 2 a du = tan a (7) 

/ esc 2 u du = — cot a (8) 

/ sec u tan u du = sec a (9) 

j esc u cot ac/a = — esc a (10) 

/tan udu= — log cos a = log sec a (n) 

/cot u du = log sin a (12) 

/sec udu = \o% (sec a 4- tan a) (13) 

/ esc u du = log (esc a — cot a) (14) 



236 Integral Calculus 

Formulae (5) to (10), inclusive, follow directly from the differ- 
ential form (§ 35, et seq.). Formulae (11) to (14), inclusive, 
may be derived as follows : 



J/ — sin u . . . 1 
tan u du = — I — — du = — log cos u = log 
J COS U ° cos u 

= log sec u. 

//'cos u . . 
cot ?/ du = I — ## = lo 
J sin u 



= log sin ^. 



//Ysec // + tan ^) sec u . 
sec // au = i au 
J sec u + tan « 

/sec 2 u du -f- tan ?/ sec z/ ^ 
sec w + tan u 

= log (sec u + tan «). 

/, Cicsc u — cot u) esc u du 
esc 11 du = I 
J cscu — cot u 



J 



esc 2 u du — esc u cot u du 



CSC u — cot u 
= log (esc u — cot u). 

EXAMPLES. 



I (sin — (-cos 2 x)dx = — 2 j — sin — 1 — I cos 2 x2dx 



1 . x 

= - sin 2^—2 cos — ■ 

2 2 



2. I sin 3 x cos ^ ^v = 



sin 4 x 



3. f^Ad$ = - fcos- 2 0(-smOdO) = - 
J cos 2 J 

= sec0, or thus, / — ^^0 = I sec tan 0d$ = sec 0. 
J cos 2 J 



4 
sin(9 jo f™-tii/ M -«/i^fl\' ( cos ^) _1 



Type Forms 237 

/sin 7. x , sec % x 
— £—dx = — 
cos 2 3 x 3 

5. I sec 2 (ax) dx = 

6. Jsec^) 



3 X 3 

tan (ax) 



«^V I ti^V - — 



a 
tan ^ 



/i -f- cos # . , , N 
; ax = log (x + sm #). 
# -+- sm a: 

8. I (tan 2 x — i) 2 dx = J [tan 2 2 x -\- \ — 2 tan 2 x\ dx 

J sec 2 2 xdx — I tan 2 x .2 dx = -^ tan 2 x -\- log cos 2 x. 

9. I (tan# + cot;r) 2 ^r = I [sec 2 x + csc 2 .r| ^r = tan^ — cot^. 

10. / (1 -f sec 2 0) 2 d0 = <9 + |tan2<9 -f log (sec 2(9 + tan2 0). 

/q / n o / n 7 tan 4 (#x) 
tan* (ax) sec (ax) dx = — - • 

/^,sin (ax) 
e <m (ax) cos (ax) dx = — : 



13. I (esc t,x -f i) 2 dx = / esc 2 T,xdx -{- 2 / esc $xdx -\- I dx 
= x — ^ cot 3 x -+- § log (esc 3 x — cot 3 x). 

sec — 

/<^x /^ dx r ~ 2 dx , x 
■= / = I = log tan-- 
sm x J . x x J x 2 2 



2 sm — cos— tan - 

22 2 



— = log tan x. 

sm x cos x 



238 Integral Calculus 

/tan x dx C sin x cos x dx 

a + b tan 2 x J a cos 2 a 



x -j- b sin 2 # 
log (a cos 2 jc 4- b sin 2 jc). 



2 (£ - a) 



C ° S2 ° dO = ( t-ism20. 



' > I + C 2 ° S2(9 ^^^+isin2^ 



. fs'm 2 0d0= C- 
. f cos 2 0d0= J 

C dQ i-n e\ _ . /tt a 

19. I - = log tan — — • Since cos ^ = sm — h 6 • 

///jc \ . a -\- b tan # 

<z 2 cos 2 x — IP 1 sin 2 .# 2 <z^ a — b tan # 

164. Type Formulae. (Continued.) Cf. §52. 

Jvftr^" 1 " (I5) 

/- vfo = ^ a (l6) 

jvf? = tan_1 " (I7) 

J-rri? = cot - 1 » (l8 > 

— . = sec -1 a (19) 

tWa 2 -* 

/ ZFT=i = csc_1 * • ( 2 °) 

C du _ x , . 

/ , = vers a (21) 



Type Forms 239 

C da _! , , 

JV=r = i ** St < 2 3) 

J ^pL= = log (« + V7±7) (24) 



Formulae (15) to (22), inclusive, may be obtained directly 
from the differential forms, § 44, et seq. Formulae (23) and 
(2 4) are derived as follows : 



/du \ C S 1 

u 2 — 1 2 J ( « — 

1 r \ du 

2 J I u — 1 



- 1 & -f- 1 
du 



du 



u + 1 ) 

= iU°g(« - i)-log(« + i)j 

« — 1 
= + 1oj 



& + I 



///z/ Cdu 



where u 2 ± 1 = .s 2 (tf) 

But from (<z), 2 #dfo = 2 s^/s ; 

//« dfe */?/ + ^2 # 
z u u -\- z 

/du Cdu -\- dz . , N 



, f ,f = log + V5^7). 



hence, 

V« 2 ± 



240 Integral Calculus 



EXAMPLES. 

du 



r du r a 

J \/a 2 — u 2 J . I u 



.11 ,« 

= sin L - or — cos -1 - 

2 a a 

2 



du 



/du i C a i . u i i 

-=— — - = - J = - tan l - or cot x - . 
a z + ir a J ir a a a a 



i + 
a* 



3. f-^==lf 

J u \lu 2 — a 2 a J 



du 



a i t u i ^u 

===== = - sec L - or esc x - 



// u 2 a 

2 



4 . /• _j^_ = /% 

J V 2 au — u 2 J I 

/du i u — a 
— - 9 = — log 
ir — a L 2 a u -\- a 

6. f , = log (// + V?^ 2 + tf 2 ). 



u lu a a a a 

V 2 _I 

<? 1 ?^ .. 7/ 

■ = vei s l - or — covers x - 



These six integrals are frequently termed auxiliary type 
forms. 



/axdx 
— 
Vl — A' 4 



^-12 

- sin i 3r 

2 



tfJC^/r /? , A 2 

8. J = - tan" 1 — 

4 + x* 4 2 



axdx a . x 2 — 2 
9 \ = _ no- 

Vr 4 - 4 8 ,0 * r 2 + 2 



Type Forms 241 

/c dx , 

= vers -1 5 x. 
Vio x — 25 a 2 

— = = 2 sin -1 y - • ^^ 
V ax — x 1 ' ' a 

I 



dx 1 ,2 IPx 

12. I = - vers 



\la 2 x - l?x 2 b a 2 



^•/ r ^=^tan-(^). 



— = —7 sec x — 



dx 1 , or 

___- — — - — —— = — sec — 

z V^ 2 * 2 - a 2 6 2 ab ab 



15. --^ - = — — x -f- tan x x. 
3 



x^fi = 

— = / ? — ■ — - = tan" 1 (a + 2). 



17 



C _ dx r 

J Va 2 — 4- A + 1 ^ «/ 



V* 2 — 4* +13 J \l(x — 2) 2 + 9 

= log (A' — 2 -f Va 2 — 41+ 13). 



*/.# 2 , 2 A + I 

= -^tan" 1 



/ • dx r 

18 ' J a 2 + x 4- 1 == 4 J 3 + (2 ^ + i) 2 " V 3 "" V3 

/dx C dx 

Va 2 + x 4- 1 J V(* +i) 2 + f 
= log \x + -J 4- V a 2 4- * 4- 1 } . 

/^a . , a* 4- 2 

— ==■ = sin -1 — — . 
V5 — 4 a — x 2 3 



242 



Integral Calculus 



21 



•/. 



dx 



J (x — 



dx 



x 2 — 2 ax cos a + a 2 J (x — a cos a) 2 + a 2 sin 9 a 



i _. x — a cos a 
- — tan L : — 



tf sin a 



a sin a 



/x 2 + x + l , 2 2a:—i 

-j — ^ ^ dfce= #+ log(^ 2 — a:+ i) + ^/= L tan -1 



23 



24 






.x 2 — x + i " 7 V3 V3 

<tfx _ 1 ( x -\- b b - i x \ 

(x 2 + a 2 ){x + b)-W+d>\ ° g V^M 2 + tf tan ~ ~a\ 



mdx 



m _ 1 2 
sec l -x. 



25 



26 



27 



■/ 

' J VI 



jr V 4 ;*: 2 — 9 3 3 

*** —sec" 1 V 5 

. — — r: See — JC. 

V5 x* - 3 ^ V3 \R 



dx 



x 2 — 4 '.x V3 

/jc 2 — 1 , 3 ^ 

-^ dx = x -f- - log - 
.ar — 4 4 # 



— ~~f log { 3 ^ — 2 + V9 x 2 — \2 x\. 



— 2 



28. 



29. 



+ 2 



> ~ *)' ^ = a sin" 1 - + Vtf 2 - x 2 . 



(a + *) 2 



/ 



^ 



30 



1 . x — a (sec a -f- tana) 

= log ; -, 

x 2 — 2 ax sec a + a 2 2 ataxia x — a (sec a — tana) 

xdx 



/2 x — 5 /* jr^ /" */.%■ 

3^ — 2 J 3 tf 2 — 2 5 J 31 2 - 

1 1 / 9 x Si X\J\ — \l2 

= _ log(3 ^_ 2) _A log _!__. 



31, 



J V7 



^jc 



3 # — ar 



. 3 4- 2 x 
= sin -1 - — — — 

V13 



Type Forms 243 

f dx i* 

32. I = tan 1 e x . 

33. I y dx = \lmx -\- x 2 -\- m log \ Vx + \lm -\- x\ . 

i 

/x -\ — 
2 
/ i' 



I I 

X -\- 
2 2 2 d 

2 4 



I ,2^+1 



= log V^ 2 H-ar + i = tan 

V3 V3 

//& 1 . 2 x — 1 + V5 



36 - /; 



I -\-X — X 2 V5 2 JC — I — V5 

dx J 1 ^ + ^ 

*** = — — = tan" 1 ——== 

a -\- 2 bx -{- ex 2 Nac — b 2 vac — b 2 



/dx 1 2 ax — b 

c + bx-ax* = ^ Sin_1 VF+" 



4 <2^r 



165. Integration by Parts. — From equation 3, § 25, of the 
Differential Calculus, we have 

d(uv) = udv -f- #d?#. 
Hence, uv = J udv + 1 ^« ; 

.*. I udv =. uv — j vdu (25) 

Examining (25) we see that the required integral is separated 
into two parts, u and dv, and that the first term of the second 
member is obtained by integrating the first member, assuming 
u constant, and that the second term is obtained from the first 



244 Integral Calculus 

term by differentiating that term, assuming v constant, and inte- 
grating the result. This process, known as integration by- 
parts, is applicable whenever dv and vdic are integrable forms. 
Let us apply the process to the example 



/< 



x sin xdx. 
Let u = x and dv = sin xdx ; then 



J'* S in^=.*(-oo.*)-/(-co8*)& 

= — x cos x + sin #. 

We might have assumed u = sin # and ^' = 37/3: ; then 

x sin •%*/.? = sin x I — cos Jc^Zr. 

2 J 2 

But the integral I — cos #/&: is more complicated than the 

given integral 1 x sin xdx ; hence this assumption will not serve 

our purpose. In applying this process, therefore, we must 
determine the proper factor by trial. 



EXAMPLES. 



1. I xlogxdx = log x 1 

= log 3: 



x 2 dx 

2 X 
2 4 



= — (log x 2 — 1). 

2. I x cos xdx = x sin x + cos x. 

3. J x sec 2 .x^r = x tan # — log sec x. 



Type Forms 245 

4. Jx^xdx=f{^x-i)xdx 



x tan x — log sec x 

6 2 



/_1 7 Jf 2 + I , # 
# tan x ^c^r = tan L x 
2 2 

6. Jsin-W* = *sin-^ + V7^. 



X* 



dx 



/x 2 i — 

a: sin -1 (xf dx ■= — sin -1 (x) 2 -\ — yi 

/sin -1 jc^t (sin -1 xf 
Vi -^ 2 ~~ 2 

/x 2 t3LW~ x xdx 
— tan -1 x (x — tan -1 x) — f (x — tan -1 x) 

J 1 -\- XT 

= x tan -1 x — (tan -1 x) 2 — log Vi + x 2 -\ — (tan -1 xf 
= x tan -1 x (tan -1 x) 2 — log Vi + x 2 . 

10. I log xdx = x (log x — i). 

- r = -e~*(x + i). 

/( dx — i \ 
xe ax dx = M 2 — J- 

C z,l ^ ** V(^ - **) 8 2 sj(a - x 2 ? 

14. I x 3 Va — x?dx = 

J 3 J 5 



246 Integral Calculus 

By repeating the process we may derive the following : 

= e^ix 2 — 2 x -f- 2). 

/£ aa: 2 JC 2 \ 

i^Afr = — (jt h -5 J. 
<z a a*/ 

17. / tf x jrV* - =-?— (x 2 - ^- + — V- J • 
J log \ log log- 0/ 

/jt 3 
a: 2 log 2 jf/atc = — (9 log 2 x — 6 log jr + 2). 
27 

19. JVlog^ 

= -. ; r^ \ (n + i) 2 log 2 X — 2 (fl + i) log * + 2 J . 

(« + 1/ 

a: 2 sin~ x xdx = — sin -1 x -\ Vi — x 2 • 

3 9 



^3 loPf ( I ~\~ X^) — X 2 

21. I x- tan" 1 xdx = — tan" 1 x + 



3 6 

22. | ^ 2 sec —1 jc^t 

.x 3 , losr (x + "v 7 ^ 2 — i") + jc V.% 2 — i 
= — sec" 1 x ^ — — — 

3 6 

e* cos - dx = ^ ( sin — h cos - ) • 

2 \ 2 2/ 

/, # sin nx — n cos /zx 
<? aa: sm ^^<w = ■ — = 5 ^. 
ar + ^ 

/_ /z sin nx -\- a cos tzjc 
«"* cos nxdx = 5— — 5 <? . 
# 2 -f- /r 



Rational Fractions 247 



CHAPTER II. 

RATIONAL FRACTIONS. 

166. The fractional differential, 

ax m + bx m ~ x + cx m ~ 2 4- . . • • + kx 4- / 
x n 4- bypf- 1 + ^ w ~ 2 + . . . . + k x x + / x 



^r . . («) 



is rational when ;;z and « are positive integers. To explain the 
method of integrating such differential forms is the object of 
this chapter.* 

167. It m> nor m = n- 

In this case we can always by division change the fractional 
form into a mixed quantity composed of one or more monomial 
terms increased or diminished by a similar fractional form, in 
which m < n. The monomial terms are readily integrated by 
rules already explained. Thus 

f4^ 3 + 2^ 2 + 4 r ( 2* + 8 ) _ 

I s— : " <WC = / < 2^ — 2 H ^ > rtfc 

J 2r+3«f2 J ( 2X? + 3X + 2 ) 

= 2/ .^^r — 2 I dx -{- 2 I 

J J J 2^r+3^+2 



* H - 4 , 
ax 



= x 2 — 2x ■+- 2 I — 7r — i — — ; — dx. 



x + 4 

2.x 2 + 3# + 2 



To complete the integration of such expression therefore we 
are to obtain a rule applicable to rational differential forms (a) 
in which m <n* 

* Leibnitz and John Bernouilli, in 1702 and 1703, showed that such integrals depended 
on the method of partial fractions. The simplified and general processes are due to Euler. 



248 Integral Calculus 

168. When m < n. If /, s, t, . . . w are the roots of the. 
equation 

then by the General Theory of Equations, 

x n + l> 1 x' l - 1 + c 1 ^-*+...k 1 x + l 1 = (x-p)(x-s)(x-t)...(x-w). 

Substituting this value in the denominator of (a) § 166, we have 

ax m -f btf*- 1 + ex m ~ 2 + . . . &* -f- / 



(x —p)(x — s)(x — t) . . . (x — a/) 



d& (p) 



By the method of Undetermined Coefficients we are enabled 
to decompose the fractional form (b) into a series of partial 
fractions of simpler forms. To do this, four cases present 
themselves, depending upon the value of the roots p,s,t, . . . w. 

Case i. When the factors of the denominator are real and 
unequal. 

Case 2. When the factors of the denominator are real and 
equal. 

Case 3. When the factors of the denominator are imaginary 
and unequal. 

Case 4. When the factors of the denominator are imaginary 
and equal. 

169. Case I. Factors real and unequal. 

Cax m + bx™- 1 + cx m ~ 2 . . . + kx +1 7 
Here I — — ax = 



/ 



(x — p) (x — s) (x — f) . . . (x — «/) 

^ ^ C ^ 

+ H + . . . \ dx = 



x — p. x — S X — / X — w 

A log (x —p) + J? log (x — s) + Clog (x — f) 4- • • • F log (x—w) 

in which A, B, C, . . . F are undetermined constants. The 
method of determining the values of the constants will appear 
in the process of integrating the following examples. 



Rational Fractions 249 



EXAMPLES. 



J X 3 — ^X 2 4" 2X J X (x — 2) (x — i) 

/ } - + | [ dx. 

J [ X X — 2 ^ — I ) 

^jc 2 — 2^: A B C 

Hence . ° — ; r = f- (- 



X {x — 2) (x — i) ^ ^ — 2 X— I ' 

clearing of fractions, we have (a) 

5X 1 — 2 x = A (x — 2)(x — 1) -f- B (x — i)# + c?(x — 2) x 
= A (x 2 - 3 # + 2) + B (x 2 - *) + C(^ 2 — 2 *) 
= (^ + ^+C)x 2 -( 3 ^+^ + 2C)jt: + 2^. 

Since the members of this equation are finite series, and the 
equation is to be satisfied for all values of x, we must have, by 
the Theory of Undetermined Coefficients, the coefficients of 
like powers of x in the two members equal ; 

i-e., a+B+C = 5, 

3A +B + 2 C= 2, 

2 A = o. 

From the three equations we find the values of these con- 
stants to be A = o, B = 4, C = — 1 ; 
hence, substituting, we have 

r 3 x 2 - 2 x ^ T c ^ r dx 

J X S — 3 X* -\- 2 X J X — 2 J X — I 

= 4 log (x — 2) — log (x — 1) 

= log (* -*y . 

& X — I 



250 Integral Calculus 

A shorter and simpler process of obtaining the values of the 
constants is as follows : 

Since (a) is true for all values of x, we may give x such 
values as will determine the constants at once : Thus 

x = o .•. o = 2 A .'. A = o 
x = i .*. i = — C .'. C = — i 

x = 2 .'. 8 = 2B .-. B = 4. 



x + 4 , 

ax = 



J 2X — X" — X* J X (i 

/! 



(i — x) (2 -J- o:) 



A B C } , 

- + + —7-- <** * 

.%• I — .T 2 + X ) 

x + 4 ^ ^ C 
Hence, — — ; — ■ r = — H H - 

X (l — X) (2 -\- X) X I — X 2 -f- X 

Clearing of fractions, we have 

x -f- 4 = A (i — x) (2 + #) -+- -^^ (2 + a:) -f- Cx (1 — ^). 
Here, jc = o gives 4 — 2^4. .*. A = 2. 

# = 1 gives 5=3^. .-. B = %. 

x = — 2 gives 2 = — 6 C. .'. C = — J. 
Hence, 

f * + 4 ^ = rh + 5 1 k 

J 2J-f-^ 3 J I X 3 (i — X) 3 (2 + X) ) 

/^ x $ r~ d x 1 r dx 
x 3 J 1 — x 3J2+X 

= 2 log x — % log (1 — ar) — J log (2 -f- •*)• 

= l°g 3 • 

V(i — #) 5 (2 + #) 



Rational Fractions 251 

_ 



3 r i 9 * + i dx= r \ a + 

J Kf + X-2 J (3^ — 1 



$x — I 5^: + 2 



^r 



= log V (3 ^ — i) 2 V(s * + 2) 3 . 

/.x 2 + 6 jp — 8 , . x*(x — 2) 
dx = log-jA-; — ^ 
x 5 — 4.x {x -f 2)- 

/•^^ _ /^ 



(# + 2) 2 



(jt — 2 + V3) (x — 2 — V3) 



2 



3 log (* - 2 - V3) - -jJ- log („ - 2 + VJ). 



2 V3 2 V3 

— dx =x + \og( — — • 

x 2 — 4 \a: + 2/ 

/x dx x 

_____ = __ 7 ^- f -6 4 log(^ + 4)-27log(^ + 3). 

r * 2 ~ 

J .x 3 — 7 



dfc 



.%• -f- 6 
- i log (* — 1) + £ log (* - 2) + r 3 o log (* + 3). 

/.T 2 + 2 JC — COS 2 a _ r ' \ I ^ 

^ 2 + 2 JC + sin 2 a t/ ( („ + i) 2 — COS 2 a ) 

, sec a , .r -f- I + COS a 

= x -\ log • 

2 x + 1 — COS a 

ia f x 2 + 8x +4. (x + i)(x — 2) 2 

10. I -3- — dx — log ^ y , — 4 • 

J ar -f- xr — 4 .r — 4 (x + 2)^ 

— = - log 

xr -\- mx m 



dx 



12 



x 2 + /#.# m x -\- m 

a -\- bx 



/dx 1 

a 2 - Px 2 = _^i g 



252 Integral Calculus 

170. Case II. Factors real and equal. 

In this case, p = s = / = ... = w. Therefore (#), § 168, 

ax m + bx m ~ x + cx m ~ 2 + . . • + kx + / , 
becomes ; ^r. 

Following the method of Case I., we would write, 

ax m 4- bx™- 1 + r^'"- 2 + • • • + kx + / , 

t v dx 

(x — p) u 

{ A B C F ) , 

= < h 1 h • • • + l dx 

( X — p x — p x — p x — p ) 

A+B+C+ . . • +F. 

= ax. 

x — p 

But this is impossible, for the given fraction cannot be re- 
duced to an equivalent fraction having a variable denominator 
(x — p) and a constant numerator (A + B + C -f- • • ■ -f- F). 
To avoid this objection, we write, 






ax m + bx m - x + ex m ~ 2 + • • • + kx + / 7 



(a: — /) n 
^ B C F ) , 



(x — p) n (x—pf- 1 (x—py- 2 (x—p) 

A B 



(i - «) (* -Z)"" 1 (2 - 71) (X -P) n ~ 2 

c 

+ ~, ^7 n 5 + ■ • • +^l0g (X — p). 

If <z// the factors of the denominator are not equal, we ascer- 
tain the partial fractions by combining Cases I. and II. 



Rational Fractions 253 



EXAMPLES. 

B C 



I _ dx = I < I 

" J (*-3) 8 J l( x ~zf 



(* — 3) 8 J l(x-3) 3 (x-3) 2 x ~3 

x 2 — 1 1 a: + 2 6 ^4 j9 C 

1 e = 1 1 ' 

(x - 3) 3 (x - 3 ) 3 (x - 3) 2 * - 3 

Clearing of fractions, we have, 

^-11x4-26=^ + ^(^-3) + C(x — 3) 2 

= Cf + ( J 5-6q^+i-3^ + 9 C. 

Equating coefficients of like powers of x, we have, 

C=i, 
B-6C=— ii, 

^ - 3^ + 9 C= 26. 

Hence, ^4 = 2 , B = — 5 , C = 1. 

/x 2 — iijc+2^ /* ( 251 

(* - 3) 3 * J U X -3T ( x ~3) 2 *- 3 

= 2 I (x — 3) -3 dfr — 5 I (x — T>)~ 2 dx -\- \ - 



dx 



dx 



x — 3 



+ -^— + log(x- 3 ). 



(* - 3) 2 x - 3 
J x* -f- 2 x l -\- x J x{x-\-i)- 

-f 



A B C I , 



(.# -f- i) 2 x +- 1 x 



254 . Integral Calculus 

It will be observed that this example affords an illustration of 
the combined methods of Cases I. and II. 

* 2 +3* + 4 A B C 

Here, — - — — = \- \- 

x z -\-2x 2 -\-x (x + i) 2 x + i x 

.-. x 2 -f- 3 x -f- 4 = Ax + Bx (x + i) + C(* + i) 2 

= (B + C)x 2 + (^ + 7? + 2 C)x+C. 

Equating coefficients of like powers, we have, 

B+ C= i, 

A + B + 2C= 3 , 

Hence, A = — 2, B = — 3, 6 = 4; 

•'• P] + 3 !s +4 ^=-2 [<*+*Y*d*-3 f— + 4 f- 
J^ 3 +2x 2 H-a: J y J 6 Jx+i*Jx 

2 <> + i) 3 

~ lQ g ..4 ' 



x -J- 1 






(^ — 2) (^ — i) 2 JC — I # — I 

3^+2 4 * + 3 1 * 2 



x,(jc -K i) 3 2 (x + i) 2 fe (# 4- 1) 2 

, C xdx _ ( x + 3 V 3 . 

J (# + 3) 2 (* + 2) g \^ + 2/ # + 3 

= log 1/ — ! 

x s — x 2 — x -\- 1 \ x— 1 2(^—1) 

- f ** 






(x 2 - i) 2 * & a; + 1 2 (x 2 - 1) 



Rational Fractions 255 



/x 5 — c x— 3 . C { 

x 2 (x + i) 2 J ( 



X 2 



$x s -{- 2x 2 — $x — 3 
! (* + i) 2 -"J (~ - ' ^(^ + i) 2 

f(i4 ^ C D \ , 

2 ^ + I < — H h - — —- - H - — > dx 

J tx" x (x -+- 1 ) a: + i ) 



rtk 



= 2 # -f ^ + \ogx(x + IT. 

2 ^r -f- # ' . 

171. Case III. Factors imaginary and unequal. 

In this case, p, s, t . . . w are imaginary and unequal. Fol- 
lowing the method of Case I., we would write, 

ax m + bx m ~ x + ^ m ~ 2 + ,■..£*+/, 
(x — p) (x — s) (x — /) • • • (x — w) 

\ A B c F \ J 

= ] + 1 + . [ dx. 

(x — p x — s x — / x — w ) 

But, as the denominators of the partial fractions are imaginary, 
the application of this process cannot afford real results. To 
develop a process applicable to this case, let us resume the 
denominator of the general fraction, (a) § 166, and equate to 
zero, i.e., 

x n + V n_1 + W n ~ 2 + • • • + k& + ^ = o . . (a) 

By hypothesis, the roots of this equation,/, s, f, etc., are im- 
aginary and unequal. We know, however, that imaginary roots 
enter equations by pairs, and that if a + b V— i is a root of (a), 
its complex conjugate, a — b V— i, is also a root. Hence, n is 
an even number. Now let p = a + b V— i, s = a — b V '— i, 
t = c + dy — i, u = c — dy 1 — i, etc. ; then, 

(* — P)( x — s ) = \* — (a -f b V— i)J {.* — (« — £ V— i)j 

= (# — tf) 2 + £ 2 , 
(# - /) (jc - u) = \x — (c + d V^7)J {# — (c — // V^i)| 

= (x — c) 2 + ^ 2 , etc. 



256 Integral Calculus 

.-. x n + V n-1 + ^ n ~ 2 + • • • + V + k 

— (x — p) (x — s) (x — t) (x — u) . . . to n factors 

= \(x - a) 2 + l?]\(x - c) 2 + <P\ . . . to^factors, 

i.e., every polynomial which, on decomposition, affords // im- 
agi?iary binomial factors of the first degree affords also - real 

2 

binomial factors of the second degree. In this case, therefore, 
we may write (V) § 1 66 in the form 






ax m + bx m ~ x + ex m ~ 2 +...+£# + / 

dx 



\{x - af + P\\(x - cf + d 2 l . . • to -factors 

2 

Ax + B Cx+£> n r . ) 

(x _^ ) + / , 2 + ( -^)^ + ' ' • to-fractions Id* . . . (b) 



in which the numerators are determined by considering, (i), 
that, in general, m = n — i , and that therefore the partial frac- 
tions, when reduced to a common denominator, must afford a 
polynomial of the same degree (m), and (2), as there are, in 
general, m -\-i (= n) terms in the numerator of the first member 
of (b) there will be m -f 1 equations of condition between the 
constants, and therefore there must he m -\-i{= n) constants, in 
order that these equations may consist. 

If, as generally happens, some of the factors of the denomi- 
nator are real and of the first degree, we combine with this 
method the methods of Case I. or of Case II., according as 
these factors are not or are repeated. 



EXAMPLES. 



C ^+2x-i r 

" J (x*+2)(x*+l) aX J 



Ax + B Cx -f- D\ , 
H 7T-. i " x - 



(X 2, -\- 2) (pC 2 -\- l) J I X 2 + 2 X 2 + I 

x 2 -f- 2 x — 1 Ax + B Cx + D 

(x 2 -\- 2) (x 2 + 1) x 2 + 2 :r 2 -f- 1 



Rational Fractions 257 

hence, clearing of fractions we have, 

x 2 -\-2x-i= (Ax + B)(x 2 + i) + (Cx + D)(x 2 + 2) 

= (^4 + 6"> 3 + (^ + D)x 2 +(A + 2 C)* +^ -h2Z>. 

Equating coefficients of like powers, we have, 

A + C= o, 

^+^= r, 
^ + 2 C = 2, 
B + 2Z>= - 1. 

.-. ^ = - 2, ^ = 3, C = 2, D = - 2 

f X 2 -\-2 X— I , /^ ( — 2 X + •* 2 X — 2 ) 7 

HeilCe ' / / 2 1 W 2 T N <** = J 2~^^ + 2 1 [ ^ 

/2 xdx f* dx C 2 xdx C dx 

X 2 4- 2 J 2 + X 2 J X 2 + I J I -\- X 2 

= — log (x 2 + 2) + -7= tan -1 — — -+- log(x 2 -f- 1) — 2 tan -1 .* 

V2 V2 

. X 2 + I ^ , ^ ! 

= log^ 1 — ^— tan -1 — : — 2 tan -1 .*. 

°jr 2 + 2 ^/ 2 y 2 

r x 2 dx r { A B Cx + D ) , 

J x (x — 1 ) t/ (a: x — 1 ^r + ^ + 1 ) 

.*. x 2 = A (x* — 1) +Bx (x 2 + x + 1) + (G* + Z>) * (x — 1) 

= (A + ^ + O x 3 + (^ - C+ D) x 2 + (^ - D)x - ^ = 0. 

.-. A +B + C= o, 

B-C + D= 1, 

B - D = o, 

^ = 0. 

... ^ = o, ^ = -,C=--,Z> = -. 

3 3 3 



258 Integral Calculus 

C x 2 dx C ( dx — \x -\- \ 1 ) 
Hence, I — r-g r = / < — -^ + -=-* -— * dx \ 

J x {x 6 — 1) J ( 3 [X — I) xr -±-x -\- I } 

3/ I C(2X— 2)dx 

= log V X — I — - I ^ 

. 3/ \ C 2 x -\- \ i /* dx 

■ log V* - - i — z / -=- — dx+- I - 

6jX 2 +X+l 2 J < 



6 J x 2 + x + i 2 J (^ + i) 2 + ? 
= log V ^ — i — 7 1°§ (^ + -^ + i) H — 7= tan x — — 

/" x 2 dx _ r I A B Cx+D i 

J (jt 2 — i) (j? 2 + 2) J ( X — I A' + I X 2 -\- 2 ) 



/X— I V2 a: 

lo g V/ 1 tan" 1 — = 

x + i 3 V2 



/.x^v . . / x 2 -f- 1 1 , 
-— — = losfV/ , -H — tan l x. 
(x + i)(^ 2 +i) te V (x + i) 2 2 

^ — dx = 3 log , + ^ tan- 1 - ■ 

^ + 4 * V* 2 + 4 22 

/rtk I (# + i) 2 I _ 1 2^— I 

x z + 1 6 >x 2 — a: + 1 -v/7 V3 

/2 x^/r A* 2 + 1 # 

/^^r 1 ( ..r . ) 

— s ^-7-= r = - < 2 tan -1 tan -1 ^ } • 
(*«+i)(*»+4) 3 ( 2 ) 

/* dx _ i /( x ~ I ) 2 * -1 •* 

' J (*_i)(**+ 2 ) ~ ° g V ^+2 " W^^ 11 V^ 



Rational Fractions 259 



x 2 cos 2 a + i 

10. f -— s ^ </* 

.ar 4- 2 ar cos 2 # + i 

sin a . x 2 -\- 2 x sin a + i cos # . 2 .%• cos <z 
log -^ = - — ■ H tan" 1 



'/; 



11. 



J x(i 



x 2 — 2 x sin a -\- 1 2 1 — jt 2 



(1 + Xj 2 (1 -f- ^ + x 2 ) 

1 



, . y/x 2 -\-X Z -\- X* I .2 X + \ 

~ + lo § / _■_ x 2 ptan" 1 — — • 

1 -f- x (1 + xf V3 V3 

/i— ^H-^ 2 , 1 _ (1 -\-x) s 1 
— . . — — — _ dx = - log —=^1 tan -1 x. 
1 + x + x 2 + x 3 2 Vi-f^ 2 2 

172. Case IV. Factors imaginary and equal. 

This case bears the same relation to Case III. that Case II. 
does to Case I. For the reasons indicated under Case II. we 
may write, therefore, equation (J?) under Case III. in the fol- 
lowing form : 



ax m + bx m ~ x + cx™- 2 + ■ ■ - + kx +/ , 






\(x- of + ^p 




j Ax + B Cx +B 


-+•■ 

- 1 


n 

■ to 


' \(x-aj + l?\* \(x — df+Pp~ 


2 









0) 

Note. — In the application of this method when the exponent of the 
denominator of a partial fraction is greater than I, the numerator being 
constant, the student will find occasion to use the following formula. 
Cf. § 186, 2°. 



/ 



x m {a + bx n ) p dx 



x" 1 * 
an 



«0+i) T an(J>+i) J K ' w 



260 Integral Calculus 



EXAMPLES. 



/V — x -f 2 _ f ( Ax +B Cx+D) 

lu J (^+i) 2 dx ~J \ C^TTJ 2 + ^+V \ dx > 

.-. * 2 — * + 2 = Ax + j5 + (Cr + D) (x 2 + i) 

= Cr 5 + Z>* 2 + (A + C) # + ^ + Z>. 

Equating coefficients, we have, 

C= o, D=i, A + C=-i, £ + £> = 2; 

hence, A = — i, B = i, C = o, D = i. 

Hence, 

J (** + i) 2 J ( O 2 + i) 2 + * 2 + i j ** 

/xdx C dx C dx 

2 +i) 2 + J (x 2 + i) 2 + J ^TT 

To integrate the last term we find on comparing it with the 
first member of the reduction formula, b, § 172, that, 

m = o,-a = 1, b = i } n = 2,J> = — 2. 

Substituting these values in the first and second members of 
(b), we have 

/(i + , 2 )-v* = - ^t^p + ^5/(1 + *r*& 

# 1 



2 (1 + f ) 2 



H — tan -1 #. 



Rational Fractions 261 



Hence 



/X 2 — X + 2 I _ X I 

"n^l — \2~ ^ = /^2 ■ \ + tan * H / o , s + - tan- 1 x 
(x? + iy 2 (xr -\- i) 2(ar + i) 2 

.%• + i 3 . 

= / o , — \ + - tan x x. 

2 (JT 4- i) 2 

J (x 2 +2) 2 ^+2 ^ V^ 

/A.xix—2) . X(-\X— i) , (x — i) 2 , , 

(x — i) 2 (x 2 +i)- {x— i)(xr-\-i) xr+i 

4 r 3* +2 ^_r 3^+2 



.-*r+ 31 



2/ 4 
1^ X — 24 26 , 2 X — ^ 

H tan" 1 



3 (x 2 - 3 x 4- 3) 3 V3 V3 

J x (a -\- bx 2 ) 2 J \ x (a + bxrf a + &r 2 ) 



1 1 , x 2 

+ — -M 



2 a (a -f- <^ 2 ) 2 # 2 a -\- bx 2 

/x*dx 1 1 1 ■ 

(1 -f jr) d .at 4- 1 4 (jr 4 i) 2 2 v y 

/^r 1 1 , x n 
= 1 log 
x (a + Ax 71 ) 2 na (a -\- bx 11 ) no 1 a + &* w 

/fl&C 1 1 

(*- i) 2 (x 2 + i) 2 = " 4(jt- i)~ 2 g ^ " ^ 

1 1 1 1 . . . . 

+ - tan x x 7 - r - — - + - log (x 2 4- 1). 

4 4 (x 2 4- 1) 4 



262 Integral Calculus 



r* 4 



+ 2^+3^ -h_3 



(** + i) 3 

2 + X J X — 8 IK 

+ Try-* v + -?r tan" 1 *. 



4<> 2 + i) 2 ' 8(** + i) 



f JC 3 + 8*+ 21 

10. / — - 



3 (* — 7) ,i, /9 N 3 Vs ,*— 2 



IRRATIONAL FRACTIONS. 

173. We have seen in the preceding discussions the methods 
of reducing rational differential fractions to one or more of the 
type integral forms, and hence the method of integrating such 
differential expressions. To integrate irrational fractional dif- 
ferentials, we have only to reduce them to a rational form, and 
then apply the methods previously developed. 

174. Methods of Rationalization. 

As irrational fractional forms occur in infinite variety, the 
method of rationalization will depend upon the particular 
form under consideration. As a general rule, however, we 
may state that the process in all cases is to substitute for the 
variable in the given expressioii a new variable that will render 
the expression rational. As illustrations of the method of ration- 
alization, we shall consider certain groups of irrational forms, 
and explain the method applicable to each group. 

175. Functions containing monomial surds only. 

Rule : Substitute for the old variable a new variable affected 
with an expone?it equal to the least common multiple of the de?iomi- 
nators of the fractional expo?ients of the old variable. 



Irrational Fractions 263 



/i i 
x? — x^ . 
1 — dx. 
x* 



EXAMPLES. 



The least common multiple of the denominators is in this 
case 12. 

Let x = z 12 , 

then x? = z 6 , x* = z 4 , x% = z*, dx = 1 2 z n dz ; 

^ — dx = I 5 — 1 2 -s 1 Vz 

= 12 f j 2 14 -2 12 | dfe 

Substituting now for z its value x&, we have 



/ 



^ X? 12 15 12 , 

— I — dx = — an* xi 

x* 15 13 



By division, of course, the result in this case could be ob- 
tained immediately. 



2. 



/ — t dx - 
J x* + 1 



Here x = z 4 , 

x* = z, x? = z 2 , dx = 4 2V3: ; 



hence / - T — l — dx = / - i — - — 4 
J x? + 1 J r + 1 



sVs 



3 



= 4/ J S 2 — — I + — ; j, \ 



i + z 1 



{ z* z 2 1 ) 

= 4 j ----*+- log (i+^+tan- 1 ;* j 



264 Integral Calculus 

/*Y*'£ T I T T T 

— - dx =4 \ -xi x? — x*-\ — losf(i+ \lx) + tan - ^1. 
X* + I (3 2 2 / 

— = -xi - - log (xi + i). 

** + i 3 3 ^ ^ ^ 

176 Fractions containing only binomial surds of the first 
degree. 

The method of rationalization is the same as in the last 
article. 



J (x 



EXAMPLES. 
dx 



(*— 3)*+ (* — 3) J 

Let x — 3 = z 6 , 

then (jc — 3)4 = s 3 , (x — 3)3 = s 2 , dx = 6 z>dz ; 

/' <& _ r 6 z^dz 

•'' o'(v-3ji + (*- 3 )rJ?T? 

J ^ + 1 



= 6 < — -f- z — log (1 + 2) > 

(3 2 ) 



Hence, substituting the value oi z ■= (x — 3)®, we have 

/^jr / 3/ 

7 v , / • — xi = 2 V*— 3-3V* -3 



4- 6V^ — 3 — 6 log (1 -f- V# — 3). 



</ 



dx 
\la + Zw 






Irrational Fractions 265 

Here a -\- bx = z 2 , 

i t~ 7 2 z dz 

Va -\- bx = z, ax = — - — • 

b 

Hence / — , = T / dz = T z =— \la -\-bx. 

J \la + bx "J o b 

3 P"~ 8 )\ ^_ 2 C (*-** '& 
J {ax - a)l ^J (x - i)* 

2 C 1 2 

= — 7= / (x — i)~*dx== —(x — i)S. 

V tf «/ 5 Vtf 

We could rationalize by making x — i = s 6 , but the process 
is obviously unnecessary in this case. 



C dx _ J__ ! \lx - 2 - A /2 

J x \lx — 2 V2 \J X _ 2 4- V2 



<■/ 



•y 



2 x — 1 



5. f v z ^ ~ x //r 



- V 



JC — A/2JC — I 



V 2 # — I . V '2 X — I — I 



^2X — I — I -\/2 X — I + I 

ix 



= 4V21-I ; + 3 10 



V(i+ \xj 

_i ( V (i+^) 3 -ym^- 3 -+ I 

128^ 3 ^ ^ V1+4* 3V(i+4^) 3 ) 



'• / 3, 



+ 1 



3 



= ;V(i + i) 2 - 3 V* + 1 + 3 log (V* + 1 + i) 3 . 



266 Integral Calculus 



/x 



= 2 Vi + x + 



V(i + xf VT+ 



X 



177. Functions containing quadratic surds of the form 

\la + bx ± x 2 . 
1. When x 2 is positive let 



\la + bx + x 2 = z — x. 
2. When x 2 is negative let 

Va + bx — x 2 = V(x — c) (d — x) = (x — c) z, 
c and d being roots of the equation, 

x 2 — bx — a = o. 



I 



EXAMPLES. 
dx 



V2 -\- $x -\- X 2 

Let V2 + 3 x ' ~r* x 2 = z — x, 

then, squaring, 2 + 3 x = z 2 — 2 zx; 

z 2 — 2 



.'. X = 



22+3 

2(£ 2 +3£+2) 

flJr = — - -^ dz, 

(2Z + 3 ) 2 

1 S 2 — 2 2 2 +32+2 

V2 + * .# + .ar = 2 — x = z — = • 

22 + 3 2 2 + 3 

Therefore, 

f ** = f-^- = log (2 . + 3) 

«^ V2+3.X + .X 2 *^ 2s + 3 

= log (2 V2 +3^ + ^ + 2^ + 3). flj 



/: 



Irrational Fractions 267 

dx 



V2 + x — X 2 
Here, . x 2 — x — 2 = (x + 1) (x — 2). 

.*. V2 + x — x 2 = \l(x -\- i)(2 — x). 

Let \l(x + i)(2 — x) = (x -\- i)z; 

then, 2 — # = (.%• -|-i)2 2 = xz 2 + z 2 . 



2 — z d 
z 2 -\- I 

dx = — rT - r- 2 rtfe, 



V2 4-* — * 2 = (*■ 4- 1) z — -=— h 1 U = L -r~ t — 

7 \2T + I / Z Z + I 

///%■ Z^* 2 ^//S" 
/ s =— I -5 = 2 cot -1 ;? 
V2 + * — X 2 J Z 2 + I 



.. 2—X 

= 2 cot -1 !/ 

V #4-1 



3. r ; dx = 2 cot-yi^. 

«7 V 2 — JV — a: 2 V # -f 2 

4. / , = log (x + Vtf 4- x 2 ). 

J \/a + x 2 

/dx . x — 1 4- Vi -\- x -\- x 2 
— = log 
jc vi -f- ^ + a; 2 a* + 1 + vi 4- x 4- .# 2 

/^r 1 , x — V2 4- v 2 — x -\- x 2 
= — - log 
x V2 — x 4- x 2 V2 x + V2" 4- V 2 — .r 4- x 2 

/dx 1 . V2 -\- 2 x — V2 — x 
= — log — - j== 
x V2 + x — x 2 V2 ' V2 4- 2 x 4- ^2 — x ' 



<1/ 



268 Integral Calculus 



dx 1 / V-* — i — x\k 

= 2 cot -1 ' °- -»• 



8. f-, ___. . r 

J V2 — 2 X — X 2 \ V3 + I + X, 

9. / = log [-+ x + \/a + bx + x 2 ) 

J ^a — bx -h- x 2 V 2 / 



rtfr I , V4 -\~ 2 X — V2 
- = — logf — — 

(2 -\- 2> x ) "^4 ~ x 2 4V2 ' V4 + 2 x + V2 



x 



11 



— =—_ log ( — -+i'Vi-+ Va+fo+r.* 2 ]' 

Na + bx + cx 2 Vc uV^ / 

/; 



<2^ 

12 



(* + *) V^ + ^ 

1 #4- ^lx 2 + b 2 + a— \la 2 +b 2 

s]a 2 +b 2 ° g *+ ^Jx 2 + b 2 +a + \I1F+T 2 



178. Methods in Special Cases. As already stated (§ 174), 
no general rule can be given for the rationalization of irrational 
forms, other than the very general rule given in that article, 
viz., to substitute in the given expression some variable that will 
effect the object in view. We shall conclude this chapter by con- 
sidering some forms as illustrative of the usual method of at- 
tacking the problem. It may be remarked that the process of 
substitution is also frequently used in simplifying irrational 
forms. 



1. 



EXAMPLES. 
x s dx 



V(^ + i) 2 
Let x 2 4- 1 = z ; 



dz dz 



then, x s = (z — i)3 ; dx — — = —, ^ , V (x 2 4- 1) 2 = si ; 

2x2 (z — 1 )* 






Irrational Fractions 269 

)dz 



V(^ 2 +i) 



— 3 



Zi 



= A S* 



4 
(* 2 + !)*(«■ -'3): 



2. Ix 3 \/a — x 2 ^jt\ 
Let a — x 2 = z 2 ; 

then, ;r 3 = (<? — z 2 )%, V<? — ^ = £, dx = — 



\la-z 2 
x 3 \Ja — .x 2 dx = — j (a — z 2 ) 2?dz = h - 

\S 3/ iS U 






J 5 



— (3 JC 2 + 2 «). 



V^ 3 — a 3 



Let *■ = •. .-.& = f^ = f^ 

13 x 2 3 yk 



2 „3 



x = y% \lx z — a s = VjT - 

Hence, / — — ^- — = f I — -J= 

«/ # V* 3 — tf 3 «/ j' Vy 2 - 



2 

— , sec" 
3«# 




2 

— -3 sec 
3#s 


"©' 



270 Integral Calculus 

dx 



I 



\f(ax 2 +b) 3 



Let x — -) .-. dx = , 

y f 



n a , yNg V(tf + ^ 2 ) 3 

Hence, f f =- f {f =- f {a + b?)-*ydy 

J V(^ 2 +^) 3 J («+£/)» J 



bVa + Jf 

x 



b \/ax 2 + £ 



/x 2 — jr 
(# — 2) 



(# — 2) 3 



We may, of course, integrate in this case by decomposing 
into partial fractions. (See § 168.) A simpler process is to 



let x — 2 = z, 

\z + 2 ) 2 - {z + 2) 



/jf 2 — .%■ 1 

I 



(X — 2) 8 J -S 3 

Z 2 + 3 £ + 2 



rtfe 



tffe 



1 3 z 

= i og 2 _ 1 _ 

-2 
= log (^ — 2) — 



6 



/ 



X — 2 (# — 2) 2 

<2& 



# (a + X s ) 



. . dz dz 

Let jc* = z. .'. rt^ = — s = t— ■ ■ 

3.x 2 3 2S 

But x = 2-5 and # + # 3 = « + z- 



Irrational Fractions 271 

C dx . C dz , CIA B \ , 

Hence ' J i&+*r v <^r v b + ^nr m 

By § 169, we find A = - and i? = ; hence, 

a a 



fz^r^ logz -l l ° 8 ^ + ^) 



= T~Z lo S 



2,a a + z 

/dx i , x 3 

— 7 ^ = — l°g 
x (a -\- x 6 ) 2> a 



x (a -\- x 3 ) 3 a a + x 3 

e*dx C dz . 

— = / -= when e* = z. 

r 1 — 4 Jr-4 

J i ^ = ilog^(by§i6 4 , 2 3). 

?^- 4 =ilog 



e* — 2. 
e* + 2 



8. J-|= = ^(3^- 4 )^ + I ) 1 



Let ^+1=2. 



dx = 

x 2 V* 4 + x 2 + i 



V^ 4 + a." 2 + i 



# 



Let jc 2 H — » = £. 



10 - f^^T^ =i( ^ + ^ i+x)i ~h(My+^' 

Let 2 + a*)i + x = z. 



272 Integral Calculus 

/**fx dx . , A c 

= | sin" 1 V- 



1. / 

Let ^2 = z. 



dx . 2 (x — b) 

12. I — — = vers -1 



J y(2 a — 



\j(2 a — x)(x — b) 2 a — b 

Let jc — b = z. 



Binomial Differentials 273 



CHAPTER III. 

BINOMIAL DIFFERENTIALS. 

179. The most general form of the binomial differential is 

x c (ax d +bx f ) h dx, (i) 

in which e, d,f, and h are any constants, positive or negative, 
entire or fractional. It is our purpose to explain the method 
of integrating these expressions. 

180. Every binomial differe?itial may be i-educed to the form 

x m (a + bx n y dx, 

in which m and n are integers, n being positive. 

For in (i), § 179, let/> d, and let us multiply and divide 
that expression by x dh . We have 

x c + dh (a + bx f ~ d fdx, 

in which f — d is a positive whole number or fraction, and 
c + dh is positive or negative, entire or fractional. 

Let us suppose these exponents are fractional, and that 

s I 

c -f- dh = ± - and / — d = + - • 
t J k 

Then the above expression takes the form 

±- l 

x t (a + bx^ h dx. 

s +sk I 

Now let x = z**, then x~* = z , x k = z lt , dx = ktz ht ~ 1 dz; 

s_ 1 

hence, x~ * (a + bx* ) h dx = ktz ' s * + kt ~ l (a + bz lr f dz, 
in which the exponents of z are integers and // positive. 



274 Integral Calculus 

Let h = p, ± sk + kt — i = m, lt — n\ then 

±- i 

" c (<zx d -f &t/) /4 dx = x * (a -j- bx& ) h dx 

= ktz± sk + u - x (a + ^)*^& 

= ktz m (a + bz n ) p dz, 

which is of the required form. We shall confine our attention 
in what follows, therefore, to the form 

x m (a -h bx n y dx, 

in which m and 11 are integers, n being positive, and/ is whole 
or fractional, positive or negative. 

RATIONALIZATION. 

181. Case I. If p is a positive integer the form 

x m (a + bx 11 ) p dx 

is rational, and may be integrated by expansion and monomial 
integration. 

Thus, I x 3 (1 + 2 x 2 ) 2 dx = I x 3 (1 + 4. x 2 -{- 4. x 4 ) dx 

= f- - x 6 + - ;r 8 . 

4 3 2 

^"/ z> # negative integer and greater than 1, we proceed as 
explained in § 172. Thus, 



f (1 -{- x 2 ) 2 dx = / q t> " i — - -h -tan 1 ^r. 



^ 1 

2 (jc 2 +1) 2 



See latter part of Ex. 1, p. 260. 



Binomial Differentials 275 

182. Case II. Where is an iiiteger or zero and p = — 

n k 

a fraction. 

In this case we have 

/ x m (a + bx n y dx = f x m (a + bx n ~fdx, 

which may be rationalized, and hence integrated by putting 

a + bx n = z 1 '. 
For under this assumption we have, 



x = 



(£ - a\h (£ - «\» , 7 * 

i —T—p x = ( —j- Y ' o + bxn ) k = ^ 



/£r = — - ( — - — ]" z k ~ 1 dz ; 
nb\ b ) 

hence, 

^ — ^v^ k Iz 1 ' — a 



I x m (a + &£")* */# = / ( 



b ) nb\ b 
k 



n z l '~ x dz 



m+l 

nb 



/m+l 
(2* -a) " V+*- 1 */*, 



772 -f- I 

- a form which is rational when is a integer or zero. 

As an illustration let us find 



I x z (l -f- 2 x 2 )^^:. 



#z -+- i z -f- I 

Here, = = 2, an integer. 

^ 2 



Therefore let 1 + 2 3? = 2 2 . 



276 Integral Calculus 



Hence, x = ( I , x z = I J , (1 + 2 x 2 )* = z % 



i\ 22 zdz 



<tx = i\ ~y- 



l\s _ V2 #//? 

Z 3 



2 . /*«*(«*- 1) 



4 V2, 



_l)Z 7 z h 



7 5 



Hence, / **(i + 2 **;!<& = J ] ^^ ^- - ^-Z ; - [ • 

J (7 5 ) 

^ tn -4~ 1 ^ 

183. Case III. Where p = — , a fraction, and h - ?> 

£ « k 



an integer or zero. 



In this case, as in Case II., we have 

This expression may be rationalized by putting 

a + bx 71 = x n z 1 '. 
For under this assumption we have, 

- V j x m = 



(a + bx n ) k = (x n z k ) k = x k z h = l-^ ) k z h , and 

# / a \z~ 1 kz k ~ 1 dz 

dx = — 



\z k - bj (z k - by' 



Binominal Differentials 277 

.-. / x m {a 4- bx n ) k dx 



-/ 



W a \ l^ a/ a \l~ x kz^dz 



z k — bj \z k — bj n \z k — b) (2* — bf 

n j \z"- — a/ (z* — b) 2 

k 
mi, 



J\z k -b) n * (z* - bj dz 



— a form which is rational and therefore integrable when 

m -j- 1 ^ . 

h - is an integer or zero. 

To illustrate this case let us write, 

/dx C 
- = I x 2 (2 + 3 x 2 ) idx. 
x 2 V2 + 3^ «/ 

m -j- 1 /z — 2 + 1 1 

Here, h - = = — 1, an integer : 

71 k 2 2 

.•.let 2 4- 3 ^ = x?z 2 . 

/ 2 \ ? 2 2 — 7. 

Hence, x = — , x~ 2 



z 2 -?,/ 2 



(2 + 3 x*)-± = (xz)-i= (*~^A 



Z--3Y1 

z 



l I 2 \ ? 4 ^//z /2r 2 — 3 \ ^ 2-^2- 

.*. \ x~ 2 (2 -\- ^ x 2 )"^ dx 

/z 2 — 3 /^ 2 — 3\* 1 /2 2 — 3\* ar^fe 
~i~' V ^ / * V^~ ) 2 - if 

/dz z m 

4 2 ' 

/dx _ j V2 4- 3 jc 2 

a^ V2 -f- 3 #* "* 



278 Integral Calculus 



EXAMPLES. 

a 2 x z . abx % 



1. fx 2 (a + bxjdx = — + — + 
J 3 3 

J v ) 7#* 5 b 2 

«/ .r 4 V i — : 



r 2 3 ^ 3 



I+a *vr=* 



/.r^Zr x 2 (x 2 ■+- 2 <? 2 ) , 

-=-: — 2 = V J + ^ 4 log V^+T 2 . 
a 2 + jc 2 4 b 

v. /**• vrn?& = iiM 8 ( 3 * 2 - 2). 

8 - J vra = — — v?^ 



3 



//JP tf 2 + 2 X 2 

^ 4 \la 2 -x 2 ~ 3 a **? 



\[rf^x 2 . 



10. f ^ = T L V(l+2X 2 ) 3 - J Vl + 2* 2 . 

/* xV.r 2 <2 4- A* 2 



(a 4- W)i p yfiT+ln* 



12 



' J (a 



Binominal Differentials 279 

o?dx x s 



(a + bx 2 )? $a(a + bx*f 



/x z dx bx 2 — 2a i —^ 
, = ™ — ya + bx 2 . 
^a + bx 2 ?>& 



14. 



15. 



/ 



\ia-\-bx 2 3^ 

dx a -\- 2 bx 2 



x? (a -\- bx 2 )? a 2 x (a -\- bx 2 )? 

dx 2 x 2 4- i 



J* 2 (i 



16 






( I + x 2 )% . x Vi + .t 2 
rtr//jt: ax 



V(i + ^) 3 VT + 



3 a: 4 — 2 .r 2 + 2 



17. / , = — V2^ + I. 

3° 



V2 ^ 2 + 1 
18. J x s (a 2 - x*)ldx = T |2 (6 x 4 - a 2 x 2 - 5 a') (a 2 - x 2 )K 



REDUCTION FORMULAE. 

184. It frequently happens that a given binomial differential 
is of such a form that the foregoing methods of rationalization 
are inapplicable. We proceed to derive a general method, by 
aid of which we are enabled to reduce any binomial differential 
to one of the type forms given in Chapter I. 

185. To deduce formula for the reduction of the exponent of the 
variable without the paretithesis. 

Two cases present themselves according as m is positive or 
negative in the form 



/■ 



x m (a + bx n ) p dx. 



280 Integral Calculus 

i. When m is positive. 

Multiplying and dividing the above form by x n ~ 1 , we have 

/*-(. + *r* -/>-*<• + »^^ 

In the formula for integration by parts, § 165, 

I ftrtfy = 7/Z' — / Vdll, 

let ?/ = x m ~ n+1 and dfr = (a + A*")**"- 1 /&-. 
Hence, 

///>(/ + 1) «*(/ + i)J v J 

x m ~ n + 1 (a + ^^ + 1 (a* — « + 1) ( /» 

«*(/ + i) ~ " nbU+i) \ J axm ~ U (* + bxn Y dx 

4- / fo m (a 4- A*")* ^ \ • 
Clearing of fractions and transposing, 

b (m — 11 4- 1) / ^ {a 4- A*")* <&: 
to the first member we have 

( „ ¥ + l!b + mb - llb + b) f^ (a + bxr , x 

(a 4- /3*") p + 1 — a(m — n+i) I x m ~ n (a 4- bx n ) p dx. 



~m — n + 1 , 

Hence, 



x m ~ n+1 (a + fo? B ) p+1 — ar(/« — «4- 1) / «"-*(« 4 bx n ) p dx 



• - 00 

6 (#/ -\-m-\-\) 



Binominal Differentials 281 

which is the required form. Formula (A), it will be observed, 
enables us to make the required integration depend upon the 
integral of an expression in which the exponent of the variable 
without the parenthesis (m) is decreased by the exponent of the 
variable within the parenthesis (n). 

2. When m is negative. 

Let m = m' + «, and therefore ni = m — n. 
Substituting these values in Formula A, and clearing of 
fractions, we have, 

= x m, + 1 (a + bx n Y + 1 - a {rri + i) j x m ' (a -f bx a y dx. 
Solving this equation for / x m '(a -\- bx n ) p dx and dropping 
accents as no longer necessary, we have, 

x m+ \a + bx n ) p + 1 — b(,ip + m + 7i-{-i) I x m+n (a + bx n ) p dx 



(m + i) 



• (Q 



This formula, commonly called Formula C, enables us to make 
the required integral depend upon another in which the expo- 
nent of the variable without the parenthesis (m) is increased 
by the exponent («) of the variable within the parenthesis. 

186. To deduce formula for the reductio?i of the exponent of the 
pare?ithesis. 

i. When p is positive. 

In this case in the general form, 



I x m (a + bof)* dx, 



282 Integral Calculus 

let u = (a + bx n ) p and dv = x m dx. Integrating by parts we 
have, 

/^ (. + **).* 

x m + 1 (a + bx n Y nbp C A , , N , , 

= - — * — / * m+w (a + b&Y- 1 dx . . . (i) 

»z + i m -\- \J ' v 7 

Applying formula (A) to the last term in the second member, 
we have, 

/*--(. + «o~* 

$(/^> + ^ 4- i) 
Substituting this value in (i), we have after reduction, 



;// + m -\- i 



• • • (*) 



Formula (i?) enables us to decrease the exponent of the 
binomial by i. 

2. When is p negative. 

Let p = p' + i and therefore, p' = p — i. 
Substituting these values in (i?), we have, after clearing of 
fractions, 

= * m+1 (0 -f- bx n Y +1 + ## (/ + i) / *"* (# + bx n Y' dx. 



Binominal Differentials 283 

Dropping accents and solving we have, 

/ x m {a + bx n ) p dx 

— x m+1 (a+bx n y +1 +(?ip + m+?i+i) f x m (a + bx n y +1 dx 

= » . (£>) 

an (J> -+- i) 

Formula (Z>) enables us to increase the exponent of the 
binomial by i. 

187. Summary. For ease of reference we collect here the 
foregoing formulae, and rearrange them in accordance with 
the letters used to designate them. 

fx m (a + bx n ) p dx 

x m - n+1 (a+bx n y +1 -a(m-n+i) Cx m ~ n {a+bx n ) p dx 

b(np+m + i) 

J x m (a + bx n y dx 

{a + bx n ) p + anp J x m (a + bx n ) p - x dx 



{A) 



x m + 



(*) 



np + m + \ 

j x m (a + bx n ) p dx 
x m+ \a+bx n ) p+1 -b(np+m + n+\)Jx M (a+bx n ) p dx 

= a(m + \) ( ° 

fx m (a + bx n ) p dx 

- x m+1 (a + bx n ) p+1 +(np+m+n + \) \ x m (a + bx n ) p+1 dx 
= ± (D) 

From an examination of the last term in the second member 
of each of these equations, we elicit the following : 

(A) decreases the exponent of the monomial factor ; 

(B) decreases the exponent of the binomial factor ; 



284 Integral Calculus 

(C) increases the exponent of the monomial factor ; 
(Z>) increases the exponent of the binomial factor. 
Or, more generally, 
{A) and (B) decrease exponents ; 
(C) and {£>) increase exponeiits. 

The terms increase and decrease are used of course in an 
algebraic sense. 

EXAMPLES. 

/x 3 dx 

In this case we observe that the given expression can be 
integrated by (i) § 162 if we can make the integral depend upon 

the integral of — — = [= x (a 2 — x?)~? dx]. This we are 
Va 2 — x 2 

enabled to do by (A) , as that formula decreases the exponent 

without the parenthesis by the exponent of the variable within 

the parenthesis. Writing, therefore, the expression in the 

general form, 

x z {a 2 — x^yi dx, 



f 



we see that m = 3, a = a 2 , b = — 1, n == 2, p = . Sub- 
stituting these values in (A) we have, 

(a 2 — x?)i — 2 a 2 J x (a 2 — x 2 )~? dx 



I x s (a 2 — x 2 ) i dx = 





X 

(a 2 - 


2 (a 2 - 


X 2 )? 


X 2 


3 

• x 2 f 


T 
2 a 2 , 




3 

! +2 


a 2 ) \ld> 


3 ( 


(V 


•■-x 2 



r 

J (a 2 — x 2 ) £ 2 xdx. 
(a 2 -x 2 f 



! - J vf 



Binominal Differentials 285 

x^dx 



Here by (A) we can make the required integration depend 

Ix 



/dx 
——=== = sin l x. In this case 
Vi 



/x^dx I 

, = / x 2 (i — x 2 )~? dx : 

Vi -x 2 J 

.'. m = 2, a = i, b = — i, « = 2, p = — ^. 

Substituting in ^4, we have, 

jc (i — x?y — I (i — ^ 2 )~i dJr 



I x 2 (i — a: 2 ) £ dk = 



/ 



— 2 

# (i — jc 2 )^ -j- sin -1 x . 



yi-^a?dx. 



we see 



Writing the example in the form I (i — x 2 )* d&e 

that the expression can be integrated if we can decrease the 
exponent of the binomial by i, thus making the integration de- 
pend upon the I (i — x 2 )~^ dx = f —= = sin -1 x. For- 
mula (B) enables us to do this. In this case m = o, a = i, 
b = — i, 7z = 2, / "= i. Substituting these values in (i?), we 
have 






2 

-1 



# / T I . 

= - Vi — xr -\ — sm —i .#. 

2 2 



286 Integral Calculus 

C dx 



3 \Jx 2 



Here / X = j x~ 3 (— a 2 + 3*)-* dk. 

In this case we wish to increase the exponent without the 
parenthesis by the exponent of the variable within the paren- 
thesis, thus making the integration depend upon 



far 1 (_ a 2 + x 2 )-? dx = f 



dx 1 , x 

= - sec J 



* Vjt 2 - a 2 a a 

Applying (C) in order to accomplish this result, making 
m = — 3, a = — a 2 , b=i,n = 2, fi=— ±, we have 

*- 2 (-tf 2 -r-:r 2 )*+ I x^^-c^ + x 2 )' 1 dx 



j x~ 3 ( — a 2 -\-x?)~*dx = 



2 a" 

yx 2 — a 2 1 , x 

-Vj 1 § sec -1 - . 

2 rtrjr 2 ar # 



rt^C 



6 - J (« 2 + * 2 ) 2 

Here f J -*L--fo + * rdx . 

Formula D is here applicable making the integration depend 

upon 

// dx 1 x 

(a> + **)- & = J j-^ = - tan- - . Making 

m = o, a = a 2 , b = 1, n = 2, j>'= — 2, 
in (Z>), we have, 

* (> 2 4- * 2 ) -1 - / (> 2 + * 2 ) -1 <& 



JW-v* 



2\-2 V^ — 



— 2 dT 

H s tan -1 



2 tf 2 (<Z 2 + ^ 2 ) 2 tf 3 tf 



Binominal Differentials 287 

In the following examples give the formulas which are appli- 
cable : 

/x dx 
. • A?is. Formula (A) twice successively. 

Vi -x> 

7> C \la 2 - x>dx ^^. Formula (^). 
J x 

/dx 
— • Ans. Formula (D). 
\/(x 2 + a 2 f 

/x 2 dx r 3 i 
= I X 1 {2 a — X) 2 <7Jt\ 
V2 ax — x 2 J 

Ans. Formula (A) twice successively. 

/dx 
7= • Ans. Formula (C). 
x 3 ya 2 — x 2 

/x^dx 
Ans. (A) once, (2?) once. 
V(« 2 — .x 2 ) 3 

u/^Vra* ^.(^) once, (*> once. 

Integrate the following : 

x 2 V# 2 — x 2 dx = — (2 x 2 — # 2 ) Va 2 — jc 2 + — sin -1 - • 
8 8 a 

14. /V^ + ??* 

4 

= ? ( 2 ^ + 5 <* 2 ) V^ + tf 2 + — log (* + V* 2 + « 2 ). 
° 8 

/<£* \//2 2 _ ~2 T ^ 
■ = _ — + — W ■ - 
x 3 \la 2 — x 2 2 a 2 .* 2 2 a z \/a 2 — x 2 -{- a 



X 



288 Integral Calculus 

— = — V2 ax — x 2 -f- a vers -1 - 

V 2 ax — x 2 a 

C x 2 dx x + -z a 1 x „ ,x 

j - = — ^— V2 ax — x 2 + ±a 2 vers -1 - 

' V2 <2^ — x 2 2 2 rt 

/x 3 d# x 2 — 4 <2 2 . 

/-y— - 2 = — ; — V*» + * 2 - 
Vr -f- « 3 

19 C dx V^ 2 - * 2 



17 

*7 V2 



20. 



/ 



°? \la 2 — x 2 a 2 x 



(a + bx*f 

x x 1 1 . jbx 2 

4- — — 7 ^ -\ ?= tan -1 1/ — 



4 £ (a + Ar 2 ) 2 8 ^ (a + ^) 8 Vtf 3 £ 3 



/xdx 1 - a , 

- = — V<^^ — x 2 -\ — vers -1 
V <?.# — x 2 2 



2 a: 



Trigonometric Integrals 289 



CHAPTER IV. 

TRIGONOMETRIC INTEGRALS. 

188. Trigonometric formulae. The following trigonometric 
relations will be found of service in what follows : 

sin 2 x -f- cos 2 x = i 

sec 2 x = i -f- tan 2 x 

esc 2 x = i + cot 2 x 
sin 2 jc = 2 sin x cos .%• 
2 sin 2 x = i — cos 2 jc 

2 COS 2 # = I -J- COS 2 X. 

189. General Rule. Separate the given expression into two 
1 factors, the first being the differential of one of the trigonometric 

functions. £xj>ress the other factor in terms of the trigonometric 
function of which the first factor is the differential. 

If this rule is followed the resulting expression is, in general, 
in an integrable form. 

For example: Jsirfx cos'*<fe. ^ 

Set aside cos xdx and observe that it is the differential of 
sin x. The remaining factor, sin 2 x cos 2 x, must now be ex- 
pressed in terms of sin x. Hence we write 



f«*x «*** =/«* * (x - sin' x) cos Xdx 



sin 2 x cos ■%•</.£ — / sin 4 x cos artfk 
sin 3 x sin 5 a: 



290 Integral Calculus 

Similarly, 

I tan^rsec 4 .*^ = / tan 3 .r (i + tan 2 x) sec 2 xdx 

= j tan 3 x sec 2 xdx + I tan 5 x sec 2 xdx 



tan 4 ^ tan 6 ^ 

+ — =— 



Or, thus 
I tan 3 a; sec 4 ^jc = / (sec 2 „r — i) sec 3 x . sec # tan xdx 

= / sec 5 jc.sec^ taxixdx — / sec z x. sextan xdx 



sec D x sec* x 



If the given expression does not contain the differential of one 
of the trigonometric functions we must attempt by aid of some 
trigonometric relation to reduce it to an equivalent expression 
that does contain such differential. Thus, 



I X.2J? xdx = I (sec 2 .* — i) dx 
= I sec 2 xdx — I , 



sec 2 xdx -■ l dx 
= tan x — x. 



Let us now examine the various trigonometric forms in a 
general way. 

190. J t&n m xdx and /cot" xdx. 

These forms can be integrated when m is an integer, positive 
or negative. For, assuming m positive, 

/C tan™ -1 .* (* 
tan m xdx= I t2in m - 2 x(sec 2 x—i)dx= I tan m - 2 xdx. 



Trigonometric Integrals 291 

The required integral is thus made to depend upon an inte- 
gral, I tan m ~ 2 xdx, in which the exponent has been diminished 

by 2. 

By repeating the process we find that the integral will ulti- 
mately depend upon / tan xdx = log sec x or / dx = x ac- 
cording as m is odd or even. 

Similarly, j o.oX? l xdx = j cot m— 2 x (csc 2 ^? — i) dx 

cot™- 1 .* r , 7 

I cot m z xdx. 



m — 



Ultimately the integral will depend upon 

Jc*** = log «n *•«/*=* 

If m is negative, then 

/ \.2J\~ m xdx = j cot m xdx, 

and I cot~ m xdx = I ta.n m xdx. 

The integration is therefore always possible when m is an 
integer. 

EXAMPLES. 

1. / tan 3 xdx = j tan x (sec 2 .* — i) dx = log sec x. 

I tan x 

2. / tan 4 xdx = tan x -\- x. 

«' 3 

/tan 4 x tan 2 ^ . . 
tan 5 xdx = h log sec x. 
4 2 



292 Integral Calculus 

/dx f 

— - — = / tan 3 xdx. Cf . i . 
cot 3 x J 

/cot 2 X 
cot 3 xdx = — log sin x. 

/'Y* "Y* "Y* 
cot 4 - dx = — cot 3 — (-3 cot \- X. 
3 3 3 



tan 5 x 



t2LV\ m + 1 X 
7)1 4- I 

COt 6 ■# 



7. I (tan 4 a* + tan 6 x) dx — 

8. / (tan'" a 4- tan" i + 2 or) dx = 

9. / (cot 5 jc 4- cot 7 x) dx = — 

10. J(car» + «*•♦•*)* = 

— = tan 4 2 tan 2 — |- 4 log sec 

,k x 4 4 



cot n+1 # 
11 4- 1 



cot°- 
4 



— —— dx = I X.23\° xdx. Cf. ^. 
cot 3 x J ° 

191. / sec M j:(£r and / csc n xdfc 

If ;z is an even positive integer these forms can be readily in- 
tegrated, for 

/f* n—2 

sec" xdx = J (tan 2 .*: 4- 1) 2 sec 2 xdx, 

csc n xdx == 1 (cot 2 a 4- 1) 2 csc? xdx, 



Trigonometric Integrals 293 

fl — 2 

in which is a positive integer. The binomial factors 

may therefore be expanded, and the terms integrated sepa- 
rately. 

When n is an odd positive i?iteger this method does not apply, as 

ft — 2 

is a fraction. For this case, see § 197. When n is a 

2 

negative integer, even or odd, we have 



For these cases see § 197. 



EXAMPLES. 



1. / sec & xdx = I (tan 2 ^ + i) 2 sec 2 xdx 

= l (tan 4 # + 2 tan 2 jt: + 1) sec 2 ^^r 



tan° x „ 

H- * tan d ^ + tan jc. 

5 



2. I csc 6 ^jc = / (cot 2 x + i) 2 csc 2 Jtv/:r 



cot 5 x 



— § cot 3 X — cot JC. 



esc 6 - dx= — 4 cot 5 2 cot 3 3 cot — 

3 3 3 3 

/„ , . ( tan 7 2 x , _ ) 

sec 8 2^^c=f< hf tan & 2^+tan d 2^H-tan 2^: > 
t 7 V 



294 Integral Calculus 

192. /tan™* sec* xdx and / cof esc" xdx. 

If n is a positive even number the method of the preceding 
article is applicable. Thus, 

/ tan 3 x sec* xdx = / tan 3 .* (tan 2 .* -f- i) sec 2 xdx 
tan 6 x tan 4 x 

If in is a positive odd number then 
tan m .* sec" xdx = / sec" -1 .* (sec 2 .* — i) 2 secx\.2j\xdx. 

cot'" a- esc" xdx = I csc* _1 * (esc 2 a — i) 2 csc .* cot a^r, 

.... m — i . ... ii- 

which since is a positive integer the binomial may be 

2 

expanded and the terms integrated separately. 

Thus / tan 3 .* sec 5 xdx = I sec 4 x (sec 2 x — i) sec x tan .%y/x 



sec' .* sec .* 



EXAMPLES. 



/sec 4 * i 

— 3— dx = I tan -4 a (tan 2 x -4- i ) sec 2 •*;/.* 
tan 4 .* J v ' y 



i i 



tan x 3 tan 3 * 

2. / tan° x sec 4 *//.* = y\ tan ¥ x + § tan s #. 

3. / tan 5 .* sec 5 xdx = T 2 T sec^ .* — ^ sec* * + § sec^ .*. 



Trigonometric Integrals 295 

4. / — 5— dx = / sec -4 x (sec 2 x — if sec x tan .xy&; 
J sec 3 a: J 



2 

= sec ^ + 



secx 3 sec 3 # 



. sec 9 ^ 2 sec 7 # sec 5 # 
tan & ^ sec 5 ^«x = 1 

9 7 5 



■1 

/ cot 5 * 



_ CSC X 
CSC JC<7JC = CSC X 

3 
_ cot 6 x cot 8 * 

CSC JC<7X = = ■ - 



8 



8. I tan s x sec 4 -x^ = § tan 2 x 



tan 1 # 



/, , . . tan 8 (jc -j- «) , 

sec 4 (.r + #) <tfr = + tan (x -f- a). 

/. . tan 3 nix tan mx 
tan 4 mxdx = — = f- .%•. 
3 #z m A 

11. I COt 5 .* CSC 5 .27fo: = — I CSC 9 .* + f CSC 7 X — \ CSC 5 .*. 



12. I ^ -^ - = | tan* * + f tan 2 #. 



'sec 4 .%y/.* 
Vcot 3 x 



193. Since 



sin x cos # 

tan .* = , cot x = — 

cos x sin x 

i i 

sec x = , esc x = — — ? 

cos x sin x 



it is obvious that all the foregoing trigonometric integrals may be 
reduced to equivalent integrals involving only sin x, or cos.*, or 
both. Let us consider now integrals involving these functions. 



296 Integral Calculus 

194. I sin m xdx and / cos'" xdx. 

i. If m is a positive odd number the integration is readily- 
performed. 

sin m jerf[r == — / (i — cos 2 ^) 2 (— sinav&r), 

cos m .*7/jc = I (i — sin 2 .*) 2 cos^v/.*:, 

. . . m — i . 

in which is a positive integer. 

Thus /*»•** ~-/(i- co**) (-<*»**) 

COS 3 a: 
= — cos x -j 

3 

= sin # — -| sin 3 jc -f- i sin 5 j?. 

2. If #2 is an even positive integer the integration may be 
affected as follows : 

// / 1 — COS 2 JtA — 
sin" 1 #//.* = / ( J 2 /&:, 

// /i -f- COS 2 Jc\ — 
COS™ .X^ = / j 1 2 ^T. 



;« 



Since — is a positive integer the binomials may be expanded 



2 



into a finite series of terms. Those terms involving odd ex- 
ponents can be integrated by the method explained under i and 
those involving even exponents may be further reduced by re- 
peating the process above. Thus 



Trigonometric Integrals 297 

/C I I — COS 2 x\ 2 
sin 4 xdx = I ( J dx 

= \ J (i — 2 COS 2 X -\- COS 2 2 *) rtfr 

x sin 2* i / r *i + cos4^ 

= 1 — I ax 

4 4 4J 2 

* sin 2* x sin 43: 
4 4 8 32 

== | x — ^ sin 2 .%■ -f- 3V sni 4 x « 

3. When #z is an even negative integer, we have 

/sin -m xdx = I — = / esc"' xdx, 
J sin™ x J 

/cos~ m xdx = I = / sec'" xdx. 
J cos"'.* J 

We proceed, therefore, as in § 191. Thus, 

/dx / cot 5 x 
-t— ~ — = I csc 6 xdx = f cot 3 * — cot*. Ex. 2, p. 203. 
sin 6 * J 5 

— s — = I sec 6 *? 7 * = -f- f tan 3 * + tan x. Ex. i,p. 203. 

cos 6 * J 5 

4. When m = — 1. 



dx 



cos- - ^sec 2 -#* 



= / = i / = i I = logtan- 

sin x J . x x J . x J x 2 

2 sin - cos - sm - tan - 

22 2 2 



x 
cos- 

2 



7T 

/dx f* dx . 2 . In x\ 
= / - = log tan = log tan • 
cos* J . Itc \ 2 \a. 2) 
sin \- - xj 



298 Integral Calculus 

5 . When m is an odd negative integer greater than i . See 
§ 197. 



>. /si 



195. / sin w ;rcos w ;ttfr. 



1. If either m or n, or both, are odd positive integers the 
method of § 194, 1, is applicable. Thus, 

1 sin 4 x cos 3 xdx = I sin 4 x (1 — sin 2 x) cos a*/.* 



sin 5 x sin 7 ^ 



/sin»*cos«*& = -/(, - «**)«*«*<- sin«*) 



cos 5 ^ cos 7 .# 



If both m and « are odd, we may adopt either of the fore- 
going methods. Thus, 



sin 4 .%■ sin 6 x 



6 ' 



or, / sin 3 x cos 3 jc^r = — 1(1 — cos 2 ^:) cos 3 ^(— sin^^r) 



cos 4 3: cos b a: 



2. When /» and 72 are even positive integers. 

Let n> m and let n =ni+p, p being an even positive 
integer. 



Trigonometric Integrals 299 

Then, / sin m #cos w .#tfk = / sin" l ^cos m+p xdx 

/*/sin 2 x\ m /i+ cos 2 x\% . 

• =J {—) ( — — ) dx 

= / sin" 1 2 x{\ -\- cos 2 x)%dx. 

2 m +?J 

p ... ... 

Since — is a positive integer the binomial can be expanded 

2 

into a series, and the terms integrated separately. 
If m > n and m = n -\-p, we have, similarly, 

/sin m .# cos* xdx = / sin n 2 x (1 — cos 2 x)* dx. 

Thus, 

J/^/sin 2 jf\ 2 
sin 4 # cos 2 #//# = I 1 J sin 2 xdx 



-s 



•/cJ«2 



sin' 2 ^ 1 — cos 2 jc\ 

due 

4 2 / 



= -| / sin 2 2 xdx —if 



I I sin 2 2 #/&: — i I sin 2 2 # cos 2 xdx 



1 i~i — COS4.X 1 sin 3 2 x 



-»/ 



d£r 



16 3 



1 sin 43: 1 . . 

= —^ # t sm d 2 x, 

16 64 48 



300 Integral Calculus 

3. When m + n is an even negative integer. 
In this case we have, 

sin'" x cos w .*#.* = I — — cos m+ n xdx 
J cos"' X 

= I tan m .* sec - (m+n) xdx. 

Since (#2 + n) is by hypothesis an even negative integer, 
— (m + n) is an even positive integer. Hence § 192, such 
trigonometric forms may be integrated. 

Thus, / 3— dx = / tan 3 x sec 2 xdx 

J COS .* J 

tan 4 .* 



EXAMPLES. 
1 . / sin 2 xdx = \ sin 2 x. 






2. / cos 2 xdx = f- J sin 2 .*. 



3. / cos 3 .r^r = sin x — ^ sin 3 .*. 



4. / s'm^xdx = — cos.* + % cos 3 .* — i cos 5 .* 






■x sin 2 .* sin 4 # 



5. I COS 4 .*^.* = — X + (- 



3 2 



6. 1 sin 5 .* cos* xdx = — \ cos 5 .* + f cos 7 .* — \ cos 9 .*. 



Trigonometric Integrals 301 

'. I sin 4 * cos 5 xdx = i sin 5 * — f sin 7 * + ^ sin 9 *. 

/dx 
i 

.. /si 



= tan * 4- h tan 3 *. 



COS" 1 * 



sin 4 xdx = — X cot 3 * — cot *. 



C 2 4 , •* sin 4* sin 3 2* 

10. I snr * cos 4 *«* = — - - 



1 6 64 48 



J 'sin* * , 1 
— #* = cos *. 
c 



1 3 
cos 2 * cos * 



„ xdx . 

12. — — — = ^ sin 5 * — 2 sin * — esc *. 



/cos 5 a 
sir 

r sin 1 
> /—I 

«/ COS' 



"\f X = I tan 1 *. 
cos 5 xdx 



dx 
14 



3 
sin 2 * cos a * 



/• dx 
. 3 3 = i tan 2 x — \ cot 2 * -f- log tan 2 *. 



snr * cos" * 



/dx /''sin 2 * + cos 2 * 
-^-= s— = / — r-^ 5 dx = tan * — cot *. 
sin" * cos' 2 * J sin- * cos" * 



196. Formulae of Reduction. 

When m and 11 are integers positive or negative, even or <?*/</, 
we can by successive reduction make the expression 



/ 



sin m * cos" xdx 



302 Integral Calculus 

depend ultimately upon an integrable form. It may be remarked 
that while the following formulae are applicable to all cases where 
the exponents are integral, yet the preceding processes should 
be employed in all cases where they are applicable as being in 
general simpler. 

197. To deduce formulae for the reduction of the expo fie fit (m) of 
sin~ X in the expression §si,f' X co? Xdx . 

i . m positive. 

I sin m x cos n xdx = — I sin m_1 .# cos".* (— s'mxdx) 

1 x I'm — i 
i J n + i 

-sm m ~ 1 x cos n + 1 x -{- {m — i) I sin m ~ 2 x cos"x (i — sin 2 x) dx 

n -(- i 

— sin" 1-1 x cos w+1 x-\- {m — \)\ sin" 1- 2 x cos w xdx — ( m — i ) / sin™* cos" xdx 



n + l 



COS 

= — sm m 1 x h / sin" 1 2 x cos n+2 xdx 

n -f- 



'l/*-*- "*<*- «-- >/"• 



n -\- l 

Clearing of fractions and transposing, we have 

(«+«)JsiD-*«*-** 

= - sin-, cos-, + (m - i)Jsfa— * «*•** 

.-. I sin™ x cos n xdx 

$ n + 1 x -\-(m — i) / sin w ~ 2 .rcos n .rdk: 



— sin m * x cos' 



m -\- n 



. . . (A) 



Trigonometric Integrals 303 

Formula A reduces the exponent (m) of sin m x by 2 . By re- 
peating the process the integration will depend ultimately upon 

I sin x cos" xdx or I cos w xdx 

according as m is odd or even. 

2. m positive and n = o. 

Making this supposition in (A), we have, 

/— sin m— 1 jccos x -f- (m — 1) / sin m_2 .x^ 
sin m A:^r = • ( B ) 
m 

3. w negative. 

Let m =— m' -\- 2, m — 2 = — #/, 

in which /#' > 2. Substituting these values in (^4), we have, 

I sin~" l ' +2 3:cos n .XY&: 

— sm~ m/+1 x cos n + 1 x -\- (— m'-\-i) I sin ~ m/ .* cos w •%•</.# 

— rri + n + 2 

Clearing of fractions, transposing, solving, and dropping ac- 
cents, we have, 

1 sm~ m x Q,os n xdx 

— sin - m+1 .a: cos w+1 ^r -\-(m — n— 2) I sin -m + 2 ^: cos n xdx 

= . A (c). 



m — 1 



By repeating the process, the integration is made to depend 
ultimately upon 

/ # cos w xdx C 
— : or I cos n xdx. 
sin x J 

according as m is odd or even. 



304 Integral Calculus 

4. m ?iegative and n = o. 
In this case (C) becomes 



/ 



sin '" x^/r 



. (Z>) 



#z — I 



198. To deduce formulae for the reduction of the exponent (n) 
of cos n x in the expression J siu m x cos' 1 xdx. 

1 . n positive. 

I sin m x cos" xdx = / cos n_1 x sin™ x cos xdx 

* x /'sin"' "*" * x 

I (« — 1) cos TO_2 ^(— sin^x) 



= cos" L X 



sin 

,M — 1 



m + 



cos' 1 1 xsin m + 1 jc n — 1 (* _ .^07 

- H I cos" J .rsin TO+ cV/m: 

w + 1 J 



7/7 -f- I 



cos n 1 xs\ri ni + 1 x 



+(«-0/^-»sin- J (.-cos^)A 

;/z + 1 
Clearing of fractions, transposing, and solving, we have 



/ 



sin™^ cos n xdx = 



-> X+ («-r)f- 



sin m+1 xcos n 1 x4-(n— 1) I sin m Jccos w 2 xdx 



{A'). 



m + n 



(A r ), after repeated use, makes the required integral depend 
upon 

/sin-. cos *&, or Jsin-^, 

according as /z is 0^ or ^?;z. 



Trigonometric Integrals 305 

2. n positive and m = o. 

(A r ) under this supposition becomes 

/sin xcos n—1 x -j- (n — i) I cos n ~ 2 xdx 
cos n xdx = • \B ) 
n 

3. n negative. 

Let n = — n' -\- 2, n — 2 = — n\ 

in which ;/> 2. Substituting in (^4') we have, 

fair***-'"** 

sin m + 1 xcos~ n/+1 x + (—n-\-i) I sin m ^ cos -n/ ^^c 
m — n r -+- 2 
Reducing and dropping accents, we have, 

f*r**x~*dx 

s ^*co S -+Lx + (n- m -2)f S m"*cos-«+>*<t* 



n — \ 



■(C) 



Repeating the application of (C), the integration will ulti- 
mately depend upon 

/"sin™* 7 r . 

I dJt:, or / sin m xtfx, 

t/ cos x J 

according as n is odd or m?/z. For further reduction we apply 
(A) or (S), § 197. 

4. « negative and m = o. 
This supposition in (C) gives 



/ sin x cos n+1 x-\-(n — 2) / cos~ n+2 ^x 
cos - " ardfc = 
n — 1 



.(//) 



306 Integral Calculus 

199. Reduction to Algebraic Forms. 

Let sin x = z ; then, 



/ sin m x cos" xdx = / z m (i — spf^dz (i) 



For x = sin 1 z. .'. dx = 



Vi - z 2 
also, cos".r = (i — sin 2 .*)* = (i — s 2 ) 5 , sin m .# = z r ' 

Since the second member of (i) is of the form, 

xT (a + bx n ) p dx, 



/• 



formulae ^4, i?, C, Z>, of §§185, 186, may be made applicable to 
trigonometric integrals. 



■ / si 



EXAMPLES. 

sin 2 x cos 2 xdx. 



The process of § 195, 2, is obviously applicable. To reduce 
the expression, however, by the reduction formula, let us apply 
(A f ), as it applies to this case. It will be observed that (A) is 
also applicable. 

In this case m = 2 and n = 2, 

/sin 3 x cos x + / sin 2 .r^r 
sin 2 x cos 2 .x^ = 
4 

sin 3 Jtr cos x H J sin 2 .# 

= (Ex. 1, p. 300). 

4 

= ^ sin 3 jf cos x -{- ^ x — -^ sin 2 #. 

2. I sin* x cos 2 xdx. 
Using (A) we have m = 4, 72 = 2. Substituting we have 



/ sin4 



x cos 2 xdx = 



Trigonometric Integrals 307 



6 
= — \ sin 3 .* cos s x + -J- (J sin 3 ;*: cos x + | jc — y 1 ^ sin 2 x) 

2 x — sin 2 jc (3 — 4 cos 2 x) sin 3 # cos jp , - N 

= — H — . (k.x. 1.) 

32 24 

3. I sec 3 Jtv/.r = J cos _3 *v&\ 



f»*x*=f< 



Using (ZX) in which « = 3, we have 



sin x cos" 



J ( 



/- ,/ COS* 

sec d *#* = 



2 COS 2 X 



+ i log tan f- - H § ^94, 4- 



1 \ 1 IT X 

= 4- < tan * sec # + log tan 

( V4 2 

4. I sec 7 xdx = ^ tan x sec # { J sec 4 x + fV sec2 x H~ f i 
+ T 5 g log (sec j? -f- tan .#). 



I esc 3 .xt/a: = I ! 



csc 3 Jtr^r = I sin 3 ^:^r 



— sin 2 .xcos.# + 

sin x 

2 

x 



= — \ cot x csc x + 1 log tan - • § 194, 4. 
csc 5 :*^ = — \ cot x csc 3 jc — I cot x csc # + f log tan - . 

C dx sin 3 x cos a: „ . 

7. I j— = I sin x cos * -f- f Jt:. 

J csc 4 * 4 8 8 



308 ntegral Calculus 

/dx cos 3 * sin x 
— j— = h | sinjc cos x 4- % x. 
sec 4 * 4 8 ¥ 

/dx 
— =— = h sec 2 * + log: tan x. 
sin * cos" 5 * Q 

/cos 4 *//* * 

— : — — = A cos** x + cos * -f- lo£ tan - . 
sin* d & 2 

/cos 4 *</* 
r-« = — i COt* (3 — COS 2 *) — %X. 
sm 2 * 2 W / 2 

J</* 
. 5 = — i cos * (esc 4 x -f- | esc 2 *) + § log tan 

Reduce the following to algebraic forms and integrate : 
13. / sin 2 * cos 2 xdx. 

Let sin * = z ; then 

dz 
sin 2 * = s 2 , cos 2 * = i — z 2 , dx 



Vi — z 2 

.*. / sin 2 * cos 2 *</* = J z 2 (i — z 2 )? dz. 
Applying formula A, § 187, we have 

x = z, m = 2, a = 1, b = — 1, n = 2, p = |-; 
hence 

s(i - * 2 )§- f(i - * 2 )*^ 

/ Z 2 (i — 2- 2 )* rtfe = 



Z (i — ^ 2 )§ I 



-|_ _ ) _ Vi — z 2 + 



1 cin-1 



sin x ar 



4/2 



= - (2 2 2 — 1) (1 — 2- 2 )* + J sin" -1 2-. 



Trigonometric Integrals 309 

Ex. 3. p. 285. Substituting for z its value, we have 



/ 



sin x ■ x 

sirr x cos^ xdx = — - — (2 sirr a: — 1) cos x + — 



sin2x 1 

= ^ sin d x cos x h — • See Lx. 1. 

16 8 

14. / — =— — 1 sec 2 jc + log tan x. 

J sin ^ cos* x 

15. I sin 2 _rd£c == \ sin 2 ^. 

/. sin 3 jc cos ^ 
sin 4 xtfx = § sin 2 jc + § x. 

/dx x 

-^— = — = — 1 cot x esc 3 x — I cot # csc # + ft log tan - . 
sm 5 x 4 8 8 & 2 

MISCELLANEOUS EXAMPLES. 

+ b sin x J / 9 x t . x\ x x 

a cos' 1 — h sin z -)-\- 2 sin - cos - 

\ 2 2/ 22 



-/• 



cos^ - 

2 



# + 2 <£ tan — \- a tan 2 - 

2 2 



-/■ 



# sec 2 - dx 

2 



a 2 + 2 ad tan- + « 2 tan 2 - + £ 2 - £ 2 

2 2 



, jc dx 
a sec* 



= 2 f U , W 

^ a 2 - s + (a tan- + ^ 

<? tan — \- b 
2 2 

- tan -1 , when a > b. 



\fa 2 -^ \la 2 -b 2 



10 Integral Calculus 

If a < b, then a 2 — ft is negative, and we write 



f f . = 2 f 

J a-\- o sin jr J 



a sec* 

2 2 



(«tan- + ^ - (ft- a 2 ) 



x 



atari- + b — \lft — 



log 



V ^-^ 2 - %tan- + £+V^=~* 



J a + o cos # J 



- 

2 



rt ( cos 2 — h sin 2 - ) + b I cos 2 sin 2 - 

2 2/ \ 2 2 



/ 



dx 



(a + £) cos 2 — h (# — ^) sin 2 - 

2 y 2 

„ see 2 - — ■ 

= 2 J U _ W 

# H- £ + (# — £) tan 2 - 

2 



<sja 2 - ft \ a + b 

H a < b, then 



V — r~7 tan ~ > when a > b. 

V a-\- b 2 



/dx C 

a -\- b cos x J 



sec 2 

2 2 



(£ — d) tan 2 - — (£ + a) 



;/: 



sec 2 

2 2 



b — a J „ x b -\- a 
tan 2 



2 b — a 
*s/b — a tan - + \jb + a 



log 



\/ft- a 2 „/t -. * 



V^ — # tan \/b + a 

2 



Trigonometric Integrals 311 

/dx , 5 tan x + * „ 
— ■ : = i tan" 1 ^ -H^ . Ex. (i). 
5 + 3 sin 2 x 4 

/</:r , , 4 tan # -f 2 
: — _ = i log ^ 
4—5 sin 2 x 



4 tan x + 8 



tan — f- 2 
2 



,. r — * — =iio g 

2 

= 4- tan" 1 i tan- V 

5 — 3 cos ^ V 2/ 



312 Integral Calculus 



CHAPTER V. 

DEFINITE INTEGRALS. 

200. Up to this point all the integrals derived have been 
indefinite, the indefinite constant of integration, C, having been 
understood to enter each integral expression. See § 160. If 
C is determi?ied from given conditions, and its value substituted 
in the integral, or if it is climiiiated altogether from the ex- 
pression, the result is termed a Definite Integral. 

201. First Method. To determine the value of C from given 
conditions. 

Let dy =f f (x) dx, 

.-. y =f 0) + Q (0 

f(x) containing, of course, no constant term. 

Now suppose the relation between x and its function y is 
such that when x = a, y = o. Substituting these values in (i), 
we have, 

o =/(*) + a 

.-. C = -/(*). 

This value of C, which is now definite in (i), gives 

y=f(x) -f(a), 
2l definite integral. 

To illustrate, let dy = x 2 dx. 

r.y = - + C. 
3 



Definite Integrals 313 

Suppose the relations of x and y are such that when x = 2, 

y = o ; then, 

o = S + C .-. C = -|. 

x 3 8 

Hence, y = 

3 3 

Again, suppose a body in a vacuum falls from a state of rest, 
and it is required to ascertain the velocity it acquires and the 
distance it falls in a time t. 

We know from mechanics that the acceleration of the velocity 
of a body falling in a vacuum is constant and = 32.2 ft. a 
second. 

Let g = 32.2 ; then, § 82, we have, 





dv 

di~ g ' 




.'. dv = gdt\ 




.-. v = gt+ C. 


But, by condition, v = 


when t = o, .*. 6 


Hence, 


z> =gt. 


Also, § 17, we have, 






ds 

M = V '' 




.'. ds = gtdt\ 




... j^^+C". 


But s = when t = ; 


.-. r=o. 


Hence, 


j = i^. 



C=o. 



Equations («), (^), give the required velocity and distance. 

202. Notation. The fact that a function (y) is zero for a 
particular value of the variable (x) that enters it is usually de- 
noted by placing the value of the variable as a subscript to the 
integral sign. Thus, 



\ 



314 Integral Calculus 

y — / f\x)dx = o 

U a 

denotes that y = o when x = a. 

203. Second Method. To eliminate C. 

Let dy =f'(x) dx; 

then, y =/{x) -f- C. (V) 

Now, suppose we know, from given conditions, that 

y = A when x = a, 
and y = B when x = £. 

Substituting these values in (#) we have, 

A=f(a)+C, 
B=f(b)+C 
Hence, ^4 - B =/(<*) -/(£). 

But since y4 and i? are values of j/, ^4 — B is also a value of y\ 
hence we may write, 

y =/(<*)-/(*)• 

This process is known as Integration between Limits, and is 
so called because it gives the value of y between certain lim- 
iting values (a and b) of x. The process eliminates C, and 
thus renders the result definite. 

204. Notation. The above process is indicated by the notation, 
j , in which a is called the superior limit and b the inferior 

limit. Thus, the expression 

y = J f(x)dx 

denotes that the integration is to be taken between the limits a 

and b of x. 



Definite Integrals 



315 



EXAMPLES. 



2-6 + C-(i-3 + 0=-|. 



1 - f (x — 3)dx = ( $x+C) 

J^dx , .x 
a ^ (-1 

Jf— \ — 

smxdx = (— cos.*) \ = — cos— + C — ( — coso-f-C) = i. 
y |- 2 

J^ dx 2 
= 2 _. 
i V^ 3 v# 

Jxctx , I — 
= log V2. 
2 I - X 2 

•p. 
f- 

Jo tf 2 

•r 






7. / 4 sec 4 ^r^c = f . 



^ 7T 

y2 + x 2 4. a 
dx Tt 



y/a 2 -*? 2 
10. I 4 sec x tan a-zj&p = V2 — 1. 

2 sin 3 x cos 3 atfJr = T J ¥ . 



12 



Jf» 00 
1 A 



V^Z7 



316 Integral Calculus 

J^ 00 dx 
o I + : 

J^ 00 dx 
o i 4- x 



13. 

XT 2 



14. / = oo. 



c 

15. I r == 7Ttf. 



V2 <7X — JT 2 



°sin 7/Jtv/jt: 1 

16. 

d'"' 4 - 2 « 



17. 



£ 



^-2Jt' COS a + i 2 sin a 



APPLICATIONS. 



In order to illustrate the foregoing methods we shall give a 
few problems to which these methods are applicable. 

1. Determine the curve whose subnormal is constant and 

equal to a. 

From § 72, we have 

dy 
Subnormal = y -=- • 
dx 

Hence, y-=- = a, 

dx 

J ydy = a J dx, 

i.e., jv 2 = 2 a (x + f). 

The required locus is therefore a parabola the position of 
whose vertex is indeterminate. If, however, we suppose y = o 
when x = o we have c = o and the equation of the curve be- 
comes 

y 2 = 2 ax. 



Definite Integrals 317 

As the position of the origin is arbitrary we may always so 
assume it as to determine c, 

2. Determine the curve whose subtangent is equal to double 
the abscissa of the point of tangency. 
From § 71 and by condition we have 

dx 



i.e 





y = 2 X, 

ay 




dy 1 dx 


•J 


y 2 x 


Integrating 


log y = h (log x -f- log c) 




= log \lcx. 


Hence, 


y — \lcx 




y 2 = ex, 



or 

i.e., a parabola with indeterminate parameter. 

3. Required the curve in which the angle between the radius 
vector and tangent is m times the vectorial angle. 

From Fig. 11 and § 77, 1, we have, since <£ = m6, 

rdO 

— — = tan ma ; 
dr 

dr cos m9 M 

— = a d6. 

r sin mu 

Integrating, log r = — (log sin mO + log c) 

r m = c sin m6. 

When m = 1 this equation represents a family of circles touch- 
ing the initial line at the pole. 

4. Required the curve whose normal is- constant and equal 
to a. From § 72 and by condition 



>\MI 



2 

= a. 



318 Integral Calculus 

Hence x 2 -f- y 1 = a 2 ; 

i.e., the circle is the required curve. The value of c is here de- 
termined as in Ex. i by assuming the position of the origin of 
coordinates. 

5. Find the locus whose polar subtangent is constant and 
equal — a. . rO = a. 

6. Find the locus whose polar subnormal is constant and 
equal a. r = ad. 

7. In the cardioid, r = a (i — cos 0), the angle between the 
tangent and radius vector is always £ the vectorial angle. 



Geometric Applications 



319 



CHAPTER VI. 



GEOMETRIC. APPLICATIONS. 

205. Definition. The process of determining the area bounded 
in whole or in part by a curve is termed Quadrature. 

206. Quadrature of plane areas. 

I. When bounded by Algebraic Curves. 

Let y =f(x) be the equation of any curve as OPC, Fig. 52, 
and let the area between the curve and the X-axis be generated 
by the ordinate (PB) of the curve moving parallel to itself from 
left to right. Let A = area 
AP'PB, and let PB be any 
position of the generating or- 
dinate y ; then dA, the incre- 
ment that A would take on 
in any unit of time provided 
the change in A became 
uniform and so continued 
throughout the unit, is evi- 
dently the area that PB would describe if its length and velocity 
remained unaltered throughout the unit. But the velocity of 
PB is the same as the rate of change of the distance OB (= x), 
i.e., it is = dx. Hence the differential area {dA) is a rectangle 
whose altitude is y {PB) and whose base is dx (BP>), i.e., 




Fig. 52. 



dA 
. A 



= ydx 



(a) 
(0 



320 Integral Calculus 

in which b and a (OB and OA) are the limits of integration 
taken along the X-axis. Equation (i) is an expression for the 
area bounded by the curve, the X-axis, and terminal ordinates. 
Similarly, we find, 

A =£xdy (2) 

the expression for the area bounded by the curve, the K-axis 
and terminal abscissae, b and a being the limits of integration 
taken along the Kaxis. 

To illustrate, let it be required to find the area of a parabolic 
segment. 

Here y 2 = 2px. .*. y = \[2~px*', 

hence, A = j ydx 

t/o 

„ /— . i ' , Vijta* _ 2 TV 

\}2pX^ClX = — 15 X S> 



-£ 



i.e., the area of any segment as OBP is § of the rectangle on 
the ordinate and abscissa, i.e., \OBPK. 

II. When bounded by Polar Curves. 

Let r =f(6) be the equation of any curve as A PC, Fig. 53, 
O being the pole and OX the initial line. Let A = OP'P, and 

let us suppose it to be generated 
by the radius vector revolving 
around O as an axis, and chan- 
ging its length in obedience to 
the law expressed in the equa- 
tion r =/(&). Obviously the rate 
Fig- 53# of change of A, i.e., dA, is the 

circular sector OPB described 
by OP in any unit of time with a constant angular velocity (d0). 
Hence, since BP = rd$ and OP = r, we have 




Geometric Applications 
dA =ir.rdO = ir 2 dO; 



.-. A 



1 



f 



'do 



321 
(*) 

(3) 



where <f> = POX and ^ = P' OX. Equation (3), it will be ob- 
served, gives the area bounded by the curve and terminal radii- 
vectores. For example, let us find the area of one loop of the 

lemniscata, 

r 2 = a 2 cos 2 6. 




Fig. 54. 



From tne equation we observe that the limiting values of 6 
are 45 and - 45 ; 



.-. = 45 and «// = - 45 . 



Hence, 



A = 



= i / r 2 dO 

IT 

c\*de=±r 



a 2 cos 2 OdO 

4 

= \ a 2 sin 2 j ; 

hence, 

i.e., the area of the loop is i- the square constructed on a. 



A = — 

2 



322 



Integral Calculus 



EXAMPLES. 



1. Find area of the circle from its rectangular equation, 

x 2 + f- = a 2 . 



Here 



ydx = 4 I V<? 5 

t/0 



x 2 dx 



{ x ,-= = a 2 . . .%• 

= 4 \ - V« — x 1 H — sin -1 - 
(2 2 a 



See Ex. 3, p. 285. 



1 « . <7 . 

= 4 < — sin -1 1 4- C sin -1 o — (7 

2 2 



4 

2 2 



7Ttf 2 . 



2. Find the area of the circle from its polar equation 
r = 2 a cos 6, the left-hand extremity of the horizontal diameter 
being the" pole and the diameter being the initial line. 



Here A = \\ r 2 d$ 






71 

^ 1 / 2 4 a 2 cos 2 6dQ 

~2 



= 2d l l- + 

I 2 



sin 2 



See Ex. 2, p. 300. 



, 7r sin 7r _ / 71 sin (— 7r) 
= 2 a 2 J - + + C - - - H i ; 



+c) j 



9 \ 7T . 7T 
= 2 AT { 1 

f 4 4 



= irar. 
r 374 5 8 



Geometric Applications 323 

3. Find the area in Ex. (2), (#), when the center is the pole 
and any diameter is the initial line ; (^), when any point on the 
curve is the pole and the tangent at that point is taken as the 
initial line. 

4. Find the area of the ellipse. -jrab. 

5. Show that the entire area of the cissoid, j? (2 a — x) = x 3 , 
is three times the area of its base circle. 

6. Find the area of the first spire generated in the spiral of 
Archimedes, r= aO. %Tr 3 a 2 . 

7. Show that the area of the cardioid, r = a (1 — cos 0), is 
six times the area of the generating circle. 

8. Area between the Witch of Agnesi, (x 2 + 4 a 2 ) y = 8 a z , 
and the ^f-axis. 4 ird 2 . 

9. Area between the cissoid, (2 a — x) y 2, = x s , and its 
asymptote, x = 2 a. 3 -na 2 . 

10. Assuming the polar equation of the cissoid, r = 2 a 
tan sin 0, and the polar equation of its asymptote, r = 2 a sec 0, 
find the area. 

11. The entire area of the hypocycloid, x* + J$ = «*, is § the 
area of its base circle. 

c , ... x = aO — a sin ) 

12.* The area of one arch of the cycloid, , m > 

y = a ( 1 — cos v) ) 

is three times the area of the rolling circle. 

ira 2 

13. Area of one loop of r = a sin 2 6. — • 

o 

ira 2 

14. Area of one loop of r = a cos nv. — ■ 

r 4 n 

* This fact was first established by Roberval in 1634. 



324 Integra] Calculus 

15. Entire area of r = a (cos 2 6 + sin 2 0). 7r# 2 . 

16. Area between y (x 2 + # 2 ) = tf 2 .* 2 and its asymptote, y — a. 

2 ^ 2 . 

17. Area between r = a (sec -\- tan 0) and its asymptote, 



r = 2 a sec 0. ( — \- 2 )a\ 

18. Area of one loop of x 2 (a 2 + _y 2 ) — y 2 (a 2 — y 2 ). 

7T — 2 



2 



19. Area between y 2 (a — x) = x 2 (a -f- x) and its asymptote, 



x = a. I — h 2 I a\ 

207. Definition. The process of determining the length of 
a curve is termed Rectification. 

208. Rectification of Curves. 
I. Algebraic Curves. 

We have found, § 18, that, 

ds = \]dx 2 + dy 2 . 

This equation may be placed in either of the following 
forms : 

*=| i+ (l)T^ 
*H-- + -(S)T* 

Hence, 5= P j i + (4|J | Vjr (i) 



Geometric Applications 325 

In (i) the limits of integration are taken along the X-axis ; in 
(2) the limits are taken along the Y-axis. 

To illustrate let us find the entire length of the circle x 2 +y 2 = a 2 . 

dy x 
From the equation we have — = Substituting in (1), 

we have /»«. f ^\ \ 



Jr x t / 
X o V 



S= (1 +- dx. 



r 

If we integrate between the limits x 1 = a and x 2 = o, we ob- 
tain the length of a quadrant ; hence for the entire length we 
have n a 1 ^\ \ f' a /a 2 \ h 

5=4 1 V + f) dx==A lXf) dx 

Jo Vtf 2 



— dx 
x 2 



Jr a dx ■ _-< x 

— = == = 4 a sin l 
v# r 



x 2 " 



7T 
= 4<2 

" 2 



= 2 na. 
II. Polaj' Curves. 
From § 76, Cor., we have 



ds= ^r*d& + dr 2 . 



Hence, ds = ),* + ( — J { &\ 

( (d6\ 2 ) i 

or, ds — { 1 + r 2 1 — - J- //r. 

H« TO 5-J ( J-' + Qj* (■) 



326 Integral Calculus 

According as the limits of integration are limiting values of 
the vectorial angle (0), or limiting values of the radius vector (r). 

To illustrate, let us find the entire length of the cardioid 
r = a (i — cos 0). 

From the equation we have, 

dr . 

— = a sin 0. 
ad 

Substituting this value together with the value of r in (i), we 

have 

r<t> 
S= I \a 2 (i - cos 0) 2 + a 2 sin 2 0|i dO. 

The limiting values of are 2 7r and o ; hence 

X2tt 
\a 2 (1 - 2 cos + cos 2 0) + a 2 sin 2 0}i ^0 

X2tt 
(1 - COS 0)* //0 

X2tt 
V2 sin 10^/0 

2^ 
= — 4 tf cos J 



= — 4 a \ cos 7r + c — (cos o + c) \ 
= 8 a. 

EXAMPLES. 

1. Find the length of the circle, the pole being at the center. 
We have for the equation of the circle 

r = # ; 
dr 



.: §208, II., (1), 5 =£" W + (o) 2 jJ 

/»2 7T 

= a J dB = 



dO 



2 ira. 



Geometric Applications 327 

2. Find the length of one arch of the cycloid. 

From the equations of the curve, Ex. 27, page 69, we 'have, 

do = asm6 > 

-^ = a(i-cosff) (a) 

dy .00 

j lm -a 2 sm - cos - a 

dy ad smO 2 2 

dx dx 1 — cos . „ 2 

m 2sm 2 - 

i </0 2 



Substituting in (1), § 208, I., the value of dx drawn from (a) 
id the \ 

we have, 



dy 
and the value of — and integrating between the limits 2 ir and o, 



S = a \ ] 1 + cot 2 - > (1 - cosO)dO 

/ 

=; 2 a I 

Jo 





'2 

= 2a I csc-sin 2 -rt# 



2ir 
sin - </0 

2 



19 

4 <? COS - 

2 



2?r 




= 8«. 



3. Find the length of an arch of a cycloid, assuming the 

rectangular form of its equation, x = a vers -1 - — V2 ay — y 2 . 

a 

dx y 

Here, 



d y ^2ay-y 2 
This value in (2), § 208, I., gives 

S = V2 a I (2 a— y)~*dy. 



*-'Vi 



328 Integral Calculus 

Integrating between the limits )\ = 2 a and y 2 = o, and 
doubling the result, we have the length of one arch ; 

Jr»2a 
(2 a — y)~%dy 


= — 4 V2 a (2 <? — jj/)a 

= 80, 

as before. See Ex. 2. 

Note. — Sir C. Wren rectified the cycloid in 1673. It was the second 
curve rectified, the semi-cubic parabola having been previously (1660) rec- 
tified by William Neil. 

4. Find the entire length of the hypocycloid, x% + y% = a*. 

dy y& 

dx x^ 

S= 4 I (1 + -S~dx 
x$l 



*s 

= 4 C& / f i 

e/0 



x %dx 



I a 
= 6 tfir^ 

lo 

= 6 a. 

a 

5. Show that the entire length of the curve, r = # sin 3 -, is 

three-fourths of the circumference of the circle of radius a. 

6. Find the length of an arc of the parabola, jy 2 = 4 ax, esti- 

\l / y t I \f rt I sy* 

mated from the origin. ^Jax + x 2 -\- a log — — • 

Na 

a I * -*\ 

7. Find the length of an arc of the catenary, y = - (e a -f- e a ) , 

estimated from the vertex of the curve. S = -[ e a - e a 



Geometric Applications 329 

P 

8. Find the length of an arc of the parabola, r — -• 

i — cos 6 

9. Find the length of the curve, 8 a z y = x* -f- 6 a 2 x 2 , measured 

x 
from the origin. - — --(x 2 4- 4# 2 )i 

o <2 

10. Find the length of the logarithmic curve, x = <?log-- 

S = a log J + y<z 2 +f + C. 

rt + V^ 2 + / 

11. Find the length of the logarithmic spiral, = a log -, 



between the limits r x and r 2 . S = V(i + ^ 2 )(^ 2 — r i)- 

12. Show that the cissoid is rectifiable. 

13. Rectify y = log between limits x 1 = 2, x 2 = i. 

S = log(<? + 



14. St. Vincent, before the middle of the seventeenth century, 
proved that any arc of the spiral of Archimedes was equal to the 
corresponding arc of a parabola. Prove it. 

209. Surfaces and Volumes of Revolution. 

Let the curve A' PC in the plane XY revolve around OX as 
an axis. Then the surface generated by the curve is a surface 
of revolution and the volume generated by the area A'CC'A is 
a volume of revolution. As every point, as P in the. curve, gen- 
erates the circumference of a circle, and every ordinate of the 
curve, as PP( = y), generates the area of a circle, we may con- 
ceive the surface and volume to be generated by the circum- 
ference and area of a circle whose center is in the X-axis 
moving in the direction of that axis, and changing its radius 
in obedience to the law expressed in the equation of the limit- 



330 



Integral Calculus 



ing curve A' PC. Assuming the latter mode of generation, let 
B be the position of the center of the generating circle at any 

instant ; then the differential or 
rate of change of the surface 
(dS) is obviously the surface 
that would be generated in any 
interval of time if PB = y re- 
mained constant throughout the 
interval, and the velocity of each 
of the generating points of the 
circumference, such as P, also re- 
mained unaltered. But such a surface is that of a cylinder 
K' PB, the circumference of whose base is 2 -k PB = 2 -ny, and 
whose altitude PK' = PK = ds = velocity of the generating 
point P of the curve A' PC. Hence 




Fig. 55- 



But 



dS = 2 iryds. 

i /dy\ 2 ) * 

ds = I 1 + I — > dx. See § 208. 
I \axj ) 

dS = 2 irv < 1 -f- [ -r- J > dx. 
( \dxj ) 



dx 



(*) 



Similarly the differential or rate of change of the volume 
(dV) is the volume that would be generated in any interval 
by the area of the circle ttPB 2, (= iry 2 ) provided its radius 
remained unaltered and the velocity of each point of that area 
also remained unaltered throughout the interval. This dif- 
ferential volume is also that of a cylinder whose base is 
it OB 2, = irf~ and whose altitude PK' = dx (not ds since each 
point of the area is moving at the instant in the direction of 
the X-axis). Hence, 



Geometric Applications 331 

dV= irf-dx (a) 

■"■ > V=tt f Xl y 2 dx (2) 

To illustrate let us find the surface and volume of the sphere 
generated by the circle, x 2 + y 2 = a 2 , revolving about the X-axis. 
(1). To find the surface. 

dy x 

Here -=- = , and y — (a 2 — x 2 )*, 

dx y v 7 



— 2 tv I y I 1 + — \ dx 
C*i a 

= 2 7T I V - 

Jxo " y 

T 

U X-2 



dx 

xz y 



= 2 ira I dx. 

Taking the limits x x = a and x 2 = o we obtain one-half of 
the surface. Hence for the whole surface we have 



S = 4 Tra J dx = 4 



(2). To find the volume. 

Integrating between the limits x 1 = a and x 2 = o and 
doubling the result we have 

V = 2 tt I y*dx 

Jo 

= 2 7r / (a 2 — x 2 ) dx 

( H \ \ a 

\ X I 

= 2 7r < ax > 

( 3 Vd 



= % Tra 3 . 



332 Integral Calculus 



EXAMPLES. 



1. Find the volume of the ellipsoid of revolution generated 

revolving tl 
(2), about Y. 



y 

by revolving the area of the ellipse, — + — = 1,(1), about X, 



4 irab 2 4 tt(i 2 1) 
Ans. , 

3 3 

2. Find the volume generated by revolving the area bounded 
by the Witch of Agnesi, (x 2 + 4a 2 ) y = 8a s , about the X-axis. 

Ans. 4 7r 2 tf 3 . 

3. Show that the volume generated by revolving the area of 
the parabola, y 2 = 2 px, about X is equal to one-half the volume 
of the circumscribing cylinder. 

4. Find the surface and volume of the cone generated by 

x y . . 

revolving the line, — f- - = 1 , and the area which it and the 
a b 

72 

axes limit, about the Jf-axis. Ans. V '= , S = 7rb \la 2 -f- P. 

3 

5. Find the volume and surface generated by revolving one 
arch of the cycloid and the area it bounds about X. 

Ans. V = 5 ttV, S = ^ ira 2 . 

6. Find the volume generated by the area of the cissoid 
{2a — x) y 2 = x s , as it revolves about its asymptote, x = 2 a. 

Ans. V = 2 7r 2 a 3 . 

7. Find the surface generated by revolving the cycloid 
about its axis. Ans. S = 8 ira 2 (w — f ). 

8. Find the volume generated by revolving the area of the 

hypocycloid, x% + y* = a%, about X. Ans. V — -tra z . 

o o 

9. Show that the volume generated by revolving one branch 
of the equilateral hyperbola, x 2 — y 2 = a 2 , about X, the limits 
of integration being x x = 2a, x 2 = a, is equivalent to sphere of 
radius a. 



Geometric Applications 



333 



10. Find the volume generated by revolving the area bounded 
by one branch of the sinusoid, y = a sin - , about X. Ans. — - • 

J 2 



11. 



Show that the expression, S = 2 tt I y\ i -\- i-j-\ > dx, 
ir J r si] 



becomes S = 2 

when the coordinates are polar 



( ST^ 



12. Find the entire surface generated by revolving the 
cardioid, r = a (1 + cos 6), about the initial line. 

Ans. S = \ 2 7ra 2 . 

210. Surfaces and Volumes in General. 

Let the surface SJBCS'p be generated by the curve SpS r 
as it moves in the direction of X, its plane being always parallel 
to ZY. 




Fig. 56. 

At the instant of reaching the position SpS f we have 
dS = area ASpS'A'. 

Let SpS' =P, SA =ds; then, since the surface is cylindrical, 

S 
dS = Pds. (a) 

.-. S= jPds. (1) 



334 Integral Calculus 

Similarly, dV '== volume ASpS'A'P/, 

== -4//#. (/;) 

.*. V = JAdx, (2) 

in which ^4 = area SJDS' and dfo = 6^. 

Formulae (1) and (2) are obviously applicable to all cases 
where P can be expressed in terms of s and where A can be 
expressed as a function of x. 

Cor. If SpS r is a circle with its center D in the X-axis ; then 

P — 2 iry, A = 7^, 

and (1) and (2) become, respectively, 

S = 2 j -nyds 

v= tt / y^c, 

as heretofore determined. See § 209. 

EXAMPLES. 

1. To find the surface and volume of a regular pyramid or 
cone. 

1. To fi?id the surface. 

Let P' — perimeter of base and Oc = // = slant height. 
Let mnd be the position of the generating perimeter P at 
any instant. Since P and P' are similar, we have 

P Od _ s . 
~P r '~~Oc~~~h'' > 



hence 




-P 


P' 




This value of P in 


(1). 


§ 210, gives 




% 




5 = 


p> nw 

^J 
P'h' 


Jdfr, 


i.e., 




S = 


2 





Geometric Applications 



335 



Hence the convex surface of any pyramid or cone is measured 
by ^ product of perimeter of its base by its slant height. 




2. To find the volume. 

Let A r = area of base and Oa = h = altitude. Let Ob = x ; 



then 



A 
~A' 



Ob 1 
Oa 



x' 
J 2 



A' 
A - p *• 



Hence, § 210, (2), V 



_A^ f h 



\A? Lt'+A' • 



V = 



A'h 



i.e., the volume of any cone or pyramid is measured by \ of 
the product of its base and altitude. 

2. Show that the volume of the frustum of any pyramid or 
h 



cone is equal to - (A + A' + \lAA') where A and A' are the 

bases, and h is its height. 

3. Find the volume of a right conoid with circular base, the 

7ta 2 h 
radius of base being a, and altitude h. Am. 

4. Find the volume of the wedges cut from a tree (radius = 
a) in cutting it down, the faces of each wedge being inclined at 
an angle of 45 . Am. %a s . 



o 



36 Integral Calculus 



CHAPTER VII. 

SUCCESSIVE INTEGRATION. 

211. Successive integration is a reversal of the process of 
successive differentiation. If, for example, x is equicrescent 
and 

d 3 y = x 2 dx 3 , 

then, J>7 = *./*■<&; 

i.e., d 2 y = dx 2 \ — \-C \ . 

U ) 
. c ,„ , r \ x z dx _ , ) 

Again, I ay = dx J < (- 6^ > ; 



,# 4 



i.e., dy = dx < 1- Cr + C, ? • 

(12 ) 

And finally, 

// \ x^dx f 

dy = I ] (- Cawk + C x dx \ ; 
J ( L2 ) 

x 5 Cx 2 
i.e., j = — h (- £> -h C/ 2 . 

In practical applications of this process the conditions given 
are, in general, sufficient to enable us to determine the values of 
C,C X . . . , and thus render the result definite. Thus, given the 
acceleration due to gravity (g), let it be required to find the 
distance s through which a body starting from rest will fall in a 
time /. 



Successive Integration 337 

By condition, 

d 2 s 

-jp=g (see § 82) ; 



ds 

~dt 



••• -. = gt + c. 



But — - = v\ and v = o when t = o, .*. C = o. 
Hence, — = p-/; 

hence, s ■= g — l-C. 

2 



But j = o when t = o, .'. C x = o ; hence, 
212. Double Integration. Let 



S = ± °-t 2 
— 2 £> l • 



x 2 y z 
u = 



6 
Differentiating (a) regarding x as constant we have 



{a) 



X 2 V 

d y u = — - dy. (b) 

Now differentiating (£), regarding y and its differential as 

constant, we have 

d x (dyti) = d 2 yor u = xy^dydx. 

A reversal of this process is Double Integration, and is indi- 
cated by using a double integral sign. Thus, if 

d 2 yx ti = xy^dydx. 
then, u = J J xf-dydx. 

In performing this operation the order of integration is 
denoted by the arrangement of the differentials proceeding 
from right to left. For example, 

u = I / x*y A dydx 



338 Integral Calculus 



=f( 



X - + CWdy 

x 4 y> Cy b „ 
+ — + C v 



20 5 

213. Triple Integration. If 

aZ z V x u — x 2 yz z dzdydx 
is the result obtained by a triple differentiation of a function of 
x, y, z, regarding only one variable to vary at a time, then 



= f f yz s dzdy\- + cl 



x*fz* Cy 2 z* Qz* „ 

2484 

and the process is termed Triple Integration. 

214. Definite Double and Triple Integrals. 

Where the limits of integration are given the result of the 
process is of course definite. Thus 



Jb Jd 



u = I I x 2 ydydx 

Jb Jd 

r a (<*-d z 

= j ydyl— — 

Jb ( 3 

c z — d s a 2 — V 1 



3 2 

<V 3 - d z ) (a 2 - F) 



6 

Jr*a r*b r*c 
I I xyzdzdydx 
Jo Jo 



b ,2 



= I I yzdzdy - 
Jo Jo 2 



Successive Integration 339 



x 



£V 2 

= / zdz 

4 

a 2 Pc 2 



It frequently happens in practical application that the limits 
of one integration are functions of the variable considered in a 
subsequent integration. Thus 



Jf*a f*a,2 — y* 
. I dydx 

o Jo 



2 ,& 
3 



a". 



EXAMPLES. 



Jf2b f*a 
I y 1 (a — x) dydx. 
b Jo 

/*« r v«2 _ X 2 
2.11 (x -\- y) dxdy. 

Jo Jo 

n>/a? - 3* 
(x 2 -j- y 2 ) dxdy. 

Jo Jo 

r* r^^zzr^. dxdy 

Jo Jo 



la cos0 

rdOdr. 



5. a 



^la 2 - x 2 

n* riiax- x * r—a— , , , 

6. f / I I dxdydz. 

Jo Jo Jo 

ra /•*yi- ? rcS 1 -^-^ 

7. 8 I / / dxdydz. 

Jo Jo Jo 



Ans. 


1 a 2 b\ 


Ans. 


%a\ 


Ans. 


7ra 4 

2 


Ans. 


ird 2 . 


Ans. 


a 2 . 


Ans. 


ira 3 . 


Ans. 


f irabc. 



340 Integral Calculus 



CHAPTER VIII. 

GEOMETRIC APPLICATIONS. 

215. Quadrature of Plane Areas. From § 206 (a), we have 

dA = ydx. 
Differentiating this equation with respect to y we have 

d 2 A = dydx. 

Hence, A = I I dydx. (1) 

From § 206, (£), we have 

dA = \ r 2 dO ; 
hence d 2 A = rdrdO ; 

: / / rdrdO (2) 



.-. A = 



The order of the differentials in (1) and (2) are obviously 
immaterial. 

Formulas (1) and (2) enable us to determine the areas 
bounded by curves by double integration. 



EXAMPLES. 



1. To find the area of the circle, x 2 + y 2 = a 2 , by double 
integration. 

In this case (1) § 215, becomes 



Geometric Applications 

= 4 / / • dxdy 

Jo Jo 

= 4 I Vtf 2 — x 2 dx 



341 



= 4 



2 2 



Ex. i, p. 322. 



~*-/y" 




= irtl . 

Fig. 58. 

2. Find the area of the circle using a polar equation. 
Let O r X, Fig. 58, be the polar axis and O the pole, then 

r = 2a cos 9 

will be its polar equation. Hence, § 215 (2), we have, 



'2,2 cos 9 



A = 



= 2 I J rdOdr 

Jo Jo 

TV 

J ^2 4 a 2 cos 2 6 
t 



i. 

•7 2 



d0 = 4 a 2 I cos 2 OdO 



= 4 # 2 



= 7T<7 2 . 



See Ex. 2, p. 322. 



3. Show that -nab measures the area of the ellipse. 

4. Find the area of the cardioid, r = a (1 — cos 0). 

Am. §7rtf 2 . 

5. Find the area between the line, ay = bx, and the parabola. 
ay 2 = bx. 

a jbx 



H* 

Ans. I a dxdy = — 

Ji>* ■ 6 b 



£ 



342 



Integral Calculus 



216. Surfaces and Volumes in General by Double and Triple 
Integration. 

I. Surfaces. From § 210 (a), we have 

dS = Pds. 

Differentiating with respect to P, we have 

d 2 S = dPds. 

Let SpS' be the position of the generating curve at any in- 
stant — p being any point of that curve. Suppose the surface 
to be generated by the curve moving in the direction of X and 

so changing that its co- 
ordinates always satisfy 
the equation of the sur- 
face,^ =/(x,z). Then, 
at the instant of reach- 
ing the position SpS' 
any point such as p has 
a motion (represented 
by pA) in a direction 
perpendicular to the 
tangent pB of the 
curve SpS' by vi?-tue of 
the motio?i of the plane of the curve in the direction of X, and 
a motion (represented by pB) in the direction of the tangent 
pB by virtue of the change in the curve as it conforms to the 
configuration of the surface, y =f(x,z), it generates ; and this 
whatever may be the absolute or resultant motion oip. But 




Fig. 59. 



Hence, 



pB = dP and pA = ds. 
d*S = dPds = area rectangle pACB. 



The projection of the rectangle pACB on the coordinate 
planes XY, XZ, YZ, are obviously dxdy, dxdz, dydz, respectively. 
If we now let <£, 6, if/ represent the angles which the plane of the 



Geometric Applications 343 

rectangle pA CB makes with XY, XZ, and YZ, respectively, we 

have 

d 2 S cos <f> = dxdy, 

d 2 S cos 6 = dxdz, 
d 2 S cos \p = dydz. 

Squaring and adding these equations, remembering that 
cos 2 <£ -J- cos 2 + cos 2 i// = i, (Ana. Geom., p. 221.) 

we have, (d 2 S) 2 = dx 2 dy 2 -f- dx 2 dz 2 + dy 2 dz 2 . 

Hence, 

<=/JHtHS7** <■> 

d z dz 

in which — and — are partial derivatives drawn from the 
dy dx 

equation of the surface, y =f(x, z). 

II. Volumes. 

From § 210 ($), we have 

dV= Adx. 
Differentiating with respect to A, we have 

d 2 V=dAdx. 
But, §206 (a), dA = ydz\ since plane SMS' is || to YZ\ 
hence, d 2 V= ydxdz. 

Differentiating with respect to y, we have 

d z V '= dxdydz ; 
Hence, 

V=j jjdxdydz (2) 



344 



Integral Calculus 



EXAMPLES. 
1. To find the surface and volume of a sphere. 

(a) To find the surface. 

Let the origin of coordinates be taken at the center of the 

sphere, and let Fig. 60 repre- 
sent the portion of the sur- 
face in the first angle, i.e., 
one-eighth of the surface. 
From the equation of the 
surface, 




Hence, 



Fig. 60. 



S=S 



x 2 + y* + z 2 = 


a\ 


we obtain, 




dz y dz 
dy z dx 


X 

z 



If' 
- *fj\ 
-If, 



, dz\ 2 fdzV\h r 7 



y 2 x 2 ^ 
1 + - 2 + -J dxdy 



a 



dxdy. 



But, 



z = Vtf 2 — x 2 — y a ; 
S=Saff- dxdy 



V 



a" — xr — y 



To determine the limits of integration we observe that X Y 
cuts from the sphere a circle whose equation is x 2 -f- y 2 = # 2 ; 
.-. y == ^la 2 — x 2 , as it everywhere measures the extension of 
Sj>S' in the direction of y, is the variable superior limit of y, its 



Geometric Appplication. 



345 



inferior limit being obviously zero. The limits in the direction 
of x are obviously x — OA (= a) and x — o. 

Hence, 

nV^rr^ dxdy 

\Ja 2 — xr — jr 

J(*a I y \ y/^~T_ 

dx ( sin 1 , 
o \ Vtf 2 - 

Jo 2 

= 4 7Ttf (x) 
/. .S = 4 7Ttf 2 . 



afo 



(d). To find the volume. 

Since s = \la 2 — x 2 — y 2 measures everywhere the extension 
of the surface in the direction of Z, we have, from § 216, (2), 

V = 8 / / / <2k//j//fc 

Jo Jo Jo 



'«V„*_. 



= 8 I f " (a 2 - a: 2 - f)\dxdy 

Jo Jo 

= 8 fVx \l^/a 2 -x 2 -y 2 -{- ( 
Jo ( 2 



2 ^2 t; 

4 -l 



V a * - ]? 



Va 2 -* 2 , 

Ex. 3, p. 285. 



2 2 



x 



3\ la 



= 2 7T I #"# 

Hence, 



V= f 7m 3 . 



346 Integral Calculus 

2. Find the entire surface of the groin formed by the inter- 
section of two equal semicircular cylinders whose radius is a. 

Assuming the axes of the cylinders as those of Z and Fwe 
have 

Jo Jo 'sja 2 — x 2 

x 2 y 2 z 2 

3. Show that the volume of the ellipsoid, -, -f —„ -h-= = i, is 

a 2 Ir c l 

two-thirds the volume of its circumscribing cylinder. 

4. Find the volume common to the two cylinders given in 
Ex. 2 . Ans. | a z . 

5. Find the volume cut from the paraboloid z 2 4- y 2 = 2 ax 
by the cylinder x 2 -j- y 2 —- ax. 

Ans. I - -h - ] a s . 
\3 4/ 



Differential Equations 347 



CHAPTER IX. 

DIFFERENTIAL EQUATIONS. 

History. — Within the last half-century the theory of ordinary differen- 
tial equations has become one of the most important branches of analysis. 

Euler's memoirs, published in 1770, gave the first method of integrating 
linear ordinary differential equations with constant coefficients. 

The science of linear partial differential equations may be said to have 
been created by Lagrange, in a series of memoirs published in 1 779-1 785, 
although Pfaff, in a paper read before the Berlin Academy in 181 5, gave 
the first general method of integrating those of the first order. 

Lie's labors in recent times have put the whole subject on a more satis- 
factory basis. 

217. Definitions. A differential equation is one which con- 
tains one or more differentials or differential coefficients. 

The general solutio?i (also called the complete integral or primi- 
tive) of a differential equation is the most general equation free 
of differentials and differential coefficients, from which the 
former may be obtained by differentiation. 

Thus, dy = 7. x 2 dx, and — = tan x, 

dx 

are differential equations, while 

y = x s + c, and y = log (c sec x), 

are their general solutions, or primitives. 

With differential equations of the above forms, or, generally, 
of the form, 

dy =/(x) dx, 

we have had to do in preceding chapters. It is our purpose in 
this to deal with some of a more general character, and to indi- 



348 Integral Calculus 

cate methods for their solution. We shall confine our attention, 
however, to ordinary differential equations, i.e., to those con- 
taining only one independent variable. 

218. Orders and Degrees. The order of a differential equa- 
tion is that of the highest differential or differential coefficient 
which enters it. 

The degree is that of the highest differential or differential 
coefficient, after the equation is freed from radicals and frac- 
tions. 

Thus, dy = 3 x 2 dx, — = tan x, 

dx 

are of the first order and first degree ; 

/dy\ 2 dy , 

-f + -f- + x = o, ds = Ndx 2 + dy 2 , 
\dxj dx 

are of the first order and second degree ; 

d 2 y dy 

— — — \- x = o, d 2 y = 6 xdx 2 , 

dx' dx 

are of the second order and first degree, and so on. 

EQUATIONS OF FIRST ORDER AND FIRST DEGREE. 

219. Form, f(x)f 1 (y)dx -f 4>(x)4>i{y)dy = 0- 

Rule: Separate the variables by division a?id i?itegr ate separately. 

EXAMPLES. 

Solve : 

1. (1 + x) ydx -f- (1 — y) xdy = o. 
Dividing through by xy, we have 

x y 



Differential Equations 349 

Integrating separately, we find 

log x + x + log y — y = c, 

or, xye x ~ y = c lf 

to be the solution. 

2. sin x cos ydx = sin y cos xdy. 

Dividing by cos y cos x, we have 

tan xdx = tan ydy. 
Hence, log sec jc = log sec y -f- log ^; 

or, sec # = <fsecj, 

is the solution. 

, \ dy o 

3. (<? — j') .%■ — — \- 2 y = o. y a xr = ce v . 

dy y 2 ~f~ i 

5. The differential relation between volume (?') and pressure 
(/) of a gas, under condition that no heat leaves or enters it 
during expansion or compression, is apdv -f- bvdp = o. Show 

a 

thatpv* = constant. 

6. A source of constant electromotive force (E) is suddenly 
introduced into a circuit of resistance, R, and self-induction, 

di 

L. The differential equation is E = Ri + L— y Show that the 

El -*?\ 

strength of the electric current (i) is i =— ( i — e l J, under the 

condition that when t = o, i = o. 

y 

7. The slope of a family of curves is ; find the equation 

of the group. xy = c. 



350 Integral Calculus 

8. Given the family of curves, y = sx ; find the curve which 
intersects each curve of the group at right angles, i.e., find their 
orthogonal trajectory, s being the variable parameter. 

9. Find the orthogonal trajectory of the hyperbolas xy — m 2 , 
7?i being the variable parameter. x 2 — y 2 = c 2 . 

220. Homogeneous Equations of form, 

f(x, y)dy+<l> (x, y) dx = 0. 
Rule : Let y = vx, a7id apply rule § 2 19. 

EXAMPLES. 

1. (x 2 — jy 2 ) dy = 2 xydx. 
Let y = vx, 

and the equation becomes 

(x 2 — v 2 x 2s ) (vdx -f- xdi!) = 2 7?x 2 dx ; 
i.e., (1 — 7?) xdv = v (1 + v 2 ) dx. 

Separating variables, we have 

1 — v 2 . dx 
dv 



v{\-\-zF) x 

.-. § 171, log v — log(i + if) + \ogc = log*; 

cr, \og—^ = \ogx. 

y 

.-. Substituting for v its value, — , and reducing, we have 

x 2 -\- f 1 = cy 
for the required solution. 



Differential Equations 351 

dv 

2. (x 2 -f y 2 ) -j- ■— xy. x 2 = 2 f log cy. 

y X X 

3. (x — 2 y) dx 4- ydy = o. log = 

c x — y 

4. x — y = V^ 2 + V 2 . jc 2 — 2 rv = r 2 . 

dx •* 

5. .arVj> + y 2 dx = — xydy. > xy 2 = c(x -\- 2 y). 

dy y - 

6. (xy — x 2 ) — = y 1 . y = «*. 

221. Form, /*(>, y ) Jy + cf> ( x , y) dx = 0, in which f( x , y) 
and <£ (j:, y) are of the first degree. 

Rule : Reduce to an homogeneous equatio?i and apply rule § 220. 

To show that such reduction is possible and to indicate the 
method of procedure let us assume the form 

(a! x -f- b'y + /) dy -f (ax -f by + c) dx = o. (1) 

Let x = x -\- m and y — y r + n ; then (1) becomes 

<w + by + ^0 <// + (W +'#/ + >&) dx' = (2) 

in which k' = dm + <£';/ + / ) / n 

>£ = am -\- bfi -\- c ) 

If now we give such values to m and n as to reduce expres- 
sions (3) to zero, equation (2) becomes 

(a'x' + £'/) a?/ + (ax* + 3/) ^ = o (4) 

which is homogeneous. 

Equating (3) to zero we find 



b'c — be' ac r — a'c 



a'b — ab' ' a'b — ab' 



(5) 



for the required values of m and n. 



352 Integral Calculus 

This process fails when 

a'b — ab' = o ; 

a' // 
i.e., when — = — . 

a b 

If, however, we let these ratios = jh j and substitute 

a' = am y , b' = bm y , 
in (i) we have 

\ m t (ax -f- by) -f- c' ' \ dy + (ax -\- by -{- c) dx = o. (6) 

Now let ax -J- by = z, 

whence, adx -f- bdy = dz. 

Substituting in (6), eliminating y and its differential we have 

dz = [ a — — - 7 )dx, 

\ m/s + c ) 

in which the variables can be separated. 

EXAMPLES. 

l. (31 - 7 y - 3) dy + (7 x — $y — 7) dx = o. 

Substituting in (5) § 221, we find 

m = 1, « = o. 
Letting y' = v' x in 

(3 ■*' - 7 2/0 ^/ + (7 *' - 3/) ^ = °> 
(cf. (4) § 221) we have, after separating variables, 

dx r 3 — 7 7/ . 
7 -T- = , 2 ^ > 

.*. log .X 77 + 2 log (V — i) + 5 log (?/ + i) "= log C. 

From x = x + ;/z, jj/ = j' + ^> since #z = 1,^ = 0, we have 

x' = x — 1 , y = y' . 

. V y 

We have also, v = — = 



x x — 1 



Differential Equations 353 

Substituting and reducing we find 

O - x + i) 2 (y + x - i) 5 = c 
to be the required equation. 

/ x dy 

2. (x-\-y — 2) — — = y — 2^: — i. 

log J2 ( 3 * - i) 2 + ( 3 y - 5 ) 2 } = V2 tan" 1 ^ * g + '■ 

3. (2 * 4- 3 y — 8) — (x + 7 — 3) — = o. 

h\og\(y-x-iy- 3 {x-iy\^\og y ~ X ~ 1 ~^~ 1 \^ = c. 

^3 y — x— i+O* — i) v3 

4. (2 .%• 4- 4 y -\- 3) dy — (2 y -\- x -\- i) dx — o. 

4 x — 8_y — log (4.x -f- 8 j + 5) = r. 

222. A Linear differential equation is an equation of the jirst 
degree with respect to the dependent variable and its derivatives. 

223. Linear Equations of the First Order. 

Form, dy 4- Py dx = Qdx- (1) 

in which ,P and <2 are functions of x. 

Rule : ./^W value of j Pdx and substitute in 

/Pdx 



ye 



/„ fPdx , . . 

Qe dx + o (2) 



• The result obtained is the required solution. 
For if we assume the form 

dy 4- Pydx = o, 

separate the variables and integrate we have 

, fPdx 

log y 4- log <? = log r, 

i.e., ye = c. (3) 



354 Integral Calculus 

If we now differentiate (3) we obtain 

/Pdx 

e {dy + Pydx) = o. 

Hence multiplying (1) by e and integrating we have (2). 

As an example let us solve 

xdy — aydx = (x + 1) dx. 

Putting in form of (1), we have 

a x-[- 1 

dy ydx = dx, 

, .. , _ a -. x -\- l 
in which P = , Q = 

/Pdx — — a I — = log — . 
J x B x a 

Substituting these values in (2) we find 

— = I — xt- dx + c. 

x 1 

.*. y = f- cx a 

1 — a a 

is the required solution. 

224. Form, dy + Pydx=Qy n dx. (1) 

Rule : Reduce to linear form and apply rule for that form. 
Cf § 223. 

This reduction may be effected by dividing through by y n and 
then letting y~ n+1 = z. The resulting equation will be linear 
in z and its derivatives. 

For example, solve 

dy ydx = =- dx. 

3 3J 



Differential Equations 355 

Here f 1 = y~ 2 . Dividing through by this we have 

ypdy y^dx = dx. 

3 3 

Now let z = yP ; dz = 3 f'dy. We have 

</z — azdx = (x -J- 1) dU:, 

which is linear in z and its derivative. Solving by the method 
of the preceding article we find 

x -}- 1 1 

for the solution. 

EXAMPLES. 

1. x (1 — x 2 ) — ax 3 = (1 — 2 ^ 2 )_y. 

dx 



y — x(a-{- c ^Ji— X s ). 

2. (ay/ + tf) //a? = (1 4- jc 2 ) </j>. j' = tfjc 4- ^ Vi + a 2 . 

3. jcy (1 + _y) //jc = (1 — x 2 ) dy. y (c Vi — x 2 + 1) = 1. 

4. <?*</)> = (1 — ^) */*:. j/r* = a: + <r. 

5. dx — .x^'Vy = xydy. x (2 — jy 2 ) H = = 1. 

6. (1 + x 2 ) dy = 2 x (2 x — y) dx. y = ± x s — x^y + c. 

dy 

7. — — \- y cos x = sin 2 x. y = 2 sin 3: — 2 + r^~ sma: . 
//# 

dy 

8. — — h y cos a: = y sin 2 x. 

dx 2 

y-» = 2 sin a; 4 h ^ (w - 1)sfalJc . 

72 — I 

9. cos xdy -f- _y sin xdx = dx. y = sin x -\- c cos a:. 



356 Integral Calculus 

dy . 
10. cos x — + sin x — i — y. y (tan x + sec x) = x -{- c. 

225. An <?.%•#<:/ differential of a function of two variables is a 
form that may result from the total differentiation of that func- 
tion. 

An exact diffe?'ential equation is an equation formed by equat- 
ing an exact differential to zero. 

Thus if u represent a function of x and y, then 

_ du du . . . 

du = — dx -\ — — ay cf . § 1 2 1 

ax ay 

du 
is an exact differential. Calling the partial derivatives — and 

— , Tl^and ^respectively, we have 
ay 

Mdx + Ndy = o (1) 

as a general form for an exact differential equation. 
As a criterion of exactness, we have (cf. § 124), 

dM _ dJV 
dy dx 

226. Exact Differential Equations. 

Form, Mdx + Ndy = 0. (1) 

Rule : Find the value of J Afdx, regarding y as coiistant ; 
substitute in the expression 

J Mdx + J(n-j- f Mdz ) d y = c (2) 

and perform the operations indicated. The result is the required 
solution. 

For if we integrate (1), we have 



/ 



Mdx + Y=c (a) 



Differential Equations 357 

in which Kis some function of y. Since (a) is a form of the 
primitive of (i) if we now differentiate with respect y, regarding 
x constant, we have 

— I Mdx 4- — = N (b) 
dy J dy 

since — = N. Solving for d Y and integrating, we find 

Y =f( N -TyS MliX ) dy {C) 

Substituting (/) in (a), we have (2). Hence the rule. 
By a method entirely analogous we may prove that 

JNdy + J(m - ^ffWy) dx=c (3) 

is the solution of (1). 

To illustrate, let us solve 

(x? + 3 xy 2 ) dx + (jy 3 + 3 o?y) dy = o. 
Here, M '= x 3 + 3 xf and N = f + 3 ^7 ; and since 

dx 6K ^ r) dy 
the equation is exact. Now applying the rule we find 

\ Mdx = f (** + 2,xf)dx = - + ^-x?f. 
■■■ S Mdx + S( N ~ lyS Mdx ) dy 

- +^x?f + / (f + $x?y - 3^)^)' = r; 



4 



42 4 



358 Integral Calculus 

is the required solution. Since the given equation is homogene- 
ous it may of course be solved by § 220. 



EXAMPLES. 

1. (x*-f)^ + x(x + 2y)=o, -+o?y - y - = c. 

2. (x 2 -{- y 2 ) dx — 2 xydy = 0. x 2 — y 2 = ex. 

- - x 

3. y (1 + ev) dx + (y — x) &dy =0. x -+- yev = c. 

4. (2 x — y -f- 1) dx + (2 j — x — 1) </y = o. 

(* + 1) (.* — y) -J- y*= <r. 

227. Integrating Factor. A factor which converts a differ- 
ential equation into an exact differential equation is called an 
integrating factor. 

That such a factor exists for every equation of the form 

Mix + Ndy = o (1) 

which admits of solutio?i is evident from the following con- 
siderations : 

If (1) is exact its primitive is of the form 

u = c. 

Cf. §225. If (1) is not exact its primitive will contain a con- 
stant of integration (e) with respect to which it may always be 
solved. If, then, we differentiate the primitive thus solved for 
e we have an exact differential equation of the form 

fi (Mdx -f- Ndy) = o (2) 

which will be satisfied for the same simultaneous values of x 
and y and their differentials as will satisfy the original inexact 
equation. Hence (2) is equivalent to that equation. Since (2) 
is of the first order and degree the factor fx may be a function 



Differential Equations 359 

of x or y, or both, but not of their derivatives or differentials. 
It is further evident, since forms of equivalent primitives are 
infinitely varied, that their exact differentials will also be infi- 
nitely varied ; hence the number of values of fx in any given 
case is unlimited. 

To illustrate : solving 2, p. 349, we have 

sec x cos y 



hence, 



sec y cos x 

cos y sin xdx — cos x sin ydy 



cos' x 

Therefore, comparing with the given differential equation, Ex. 
2, p. 349, we see that — ^— is an integrating factor. We see also 

that , the factor by aid of which the variables were 

cos x cos y 

separated, is another value of /x. 

Again, in Ex. 1, p. 348, we see that - — is an integrating 
factor. Differentiating 

xye x ~ y = c, 
we have, xye c ~ v idx — dy) -j- e x ~ y {xdy -\- ydx) = o . 
i.e., e*~ y 5(i + x) ydx + (1 — y) xdy\ = o. 

Hence e*^ is another value of /x in this case. 

228. The following methods may be observed in determining 
a value of /x in certain cases. 

1. By inspection. 

Thus if the equation, 

x (1 — y 3 ) dy 4- ydx = o, 

is placed in the form, 

xdy + ydx — xjfdy = o, 



360 Integral Calculus 



we readily see that — is an integrating factor and that 

f 

log xy = c 

is the solution. 

i 



2. If (i) is homogeneous ix = . 

J v ' * r Mx + Ny 

Take, for example, the equation, 

(x 2 — y 2 ) dx + xydy = o. (a) 

In this case, Mx + Ny = x s — xy 2 -f- xy 2 = x s ; hence 

i i 



Mx + Ny x 3 

Multiplying (a) by this factor, we have 

x 2 — y 2 y 

— ~» — «•* + — 2 ^ = ° ; 

which is £%•#<:/, since 

,/y _ *» _ dx ' Ct * 225 ' (2> 
Hence, applying § 226, (2), we have 

y 2 

log x H 5 = *• 

2r 

for the solution. 

If Mx -\-Ny=o, this method fails, but the solution is y = £r. 

3. Jf (f) is of the form 

<f>(x,y)ydx + ^ 0,3;)* <fy = 0, 

r Mx — Ny 
Take as an example, 

(1 -f- xy) ydx — (xy — 1) xdy = o. (£) 



Differential Equations 361 



Here, Mix — Ny = xy + x^y 2, + x 2 f" — xy = 2 ^y 2 ; 

1 1 



.M* — Ny 2 ^y 2 

Multiplying (b) by this factor, we have 

1 + xy xy — 1 

tf.r — ay = o. 



Since 



2 jt>> 2 j' jc 

5 M 1 dilT 



^/y 2 x^y 2 dx 

the equation is exact. Applying § 226, (2), we find, 

x i 
- = a? 1 ' 

to be the solution. 

If Mx —Ny = o, this method fails, but the solution is xy = c. 

■ 1 /dM dN\ , . , /*<*)** 

1 /3iV BM\ , N , /*<*)<*> 

° r ' lf M^"^/^ 00 ' ^ = ' ' 

Take for example, the equation, 

(x 2 + j 2 ) dx — 2 xydy =■ o. (V) 

1 /aj/ aiV\ 2 
Here ' ^(^r-^j = -^ = ^)' 

f$(x)dx f—- x c1x l °S-^ I 

Multiplying (V) by this value, we have 

** + f , y , 

-5 — ^r — 2 - #y = o, 

which is exact, since 

dM _2y _dN 
dy x 2 dx 



362 Integral Calculus 

Applying § 226, (2), we find the solution to be 

x 2 — y 2, = ex. 

EXAMPLES. 

X ex* 

1. (x*-y — 2 xy 2 ) dx + (3 x 2 y — x 3 ) dy — o. - — log — • 

y 

2. (x^y 2 + xy) ydx + (x 2 y 2 — 1 ) xdy = o. xy = log — • 

3. x 2 -\- 2xy — y 2 = (x 2 — 2xy — jy 2 ) — • x 2 -\- y 2 = c(x-\- y). 

4. x 2 -f- 2 x + y 2 -f- 2 y — - = o. ^(jk 2 + y 2 ) = ^- 

dx ' 

5 . 2 xydx + (j 3 — 3 .a- 2 ) ^/y = o . x 2 + jy 3 log j = ry 3 . 



EQUATIONS OF FIRST ORDER AND A^ TH DEGREE. 

229. Form, f ( p \ p «~\ . . . x , y) = (1) 

m which p = — . 
dx 

1. When the first member of (1) <r<2^ <fe resolved into 11 rational 

bino??iial factors of the first degree with respect to p. 

Rule : Factor, equate each factor to zero and solve separately. 

Take for example the equation 



(dy\ 2 . N dy 



dy 
Letting / = — - , transposing and factoring we have 

(p -f- x) (p - y) = o. 

Equating each factor to zero and solving we find 

2 y + x 2 — c = o, 
and y = ce*. 



Differential Equations 363 

Hence, (2 y -f- x 2 — c) (y — ce*) = o 

is the general solution. 

2. When (1) can be put in the form 

y = f(p,x) (<*) 

Rule : Differentiate (a), thus obtaining 

#« equation of the first order between p and x. Solve (b), and 
eliminate p between the resulting equation and (a). 

3. When (1) can be put in the form 

* = f(p> y) W 

Rule : Differentiate (V) thus obtaining 

dx 
in which p = — . Solve (d), and elimifiate p between the result- 
ing equation and (c). 

Take for example the equation 

y=*f+f. 
Differentiating we have 

i.e., /(i-2(/ + i)|) = o. 

Hence, p = o, 

and 2 (p + 1) dp = dx. 

Integrating the latter we find 

p 2 -}- 2 p = x + c; 
hence, / = V# -f- c — 1 . 



364 Integral Calculus 

This value of p in the given equation gives, after reducing 
and rationalizing, 

(3 x + 31 + 3 c ~ J ) 2 = 4 (x + cf 

for the general solution. The first value, p = o, when substi- 
tuted in the given equation gives the singular solution y = o. 

The success of this method depends (1) on our ability to solve 
the derived equations (J?) or (d), and (2) upon our ability to 
eliminate p. 

The method can in general be successfully employed in 
equations in which one of the variables is absent, and in those 
where both enter but are of the first degree. 

EXAMPLES. 

dv" 

1. -~ = a 2 y 2 . (y— ^0 (v — ee~ ax ) = o. 
dor 

2. p 2 — §p -\- 6 = o. ( y — 2 x — c) (y — 3 x - c) = o. 

3. p 2 — 3P + 2 = 0. ( y — 2 x — c) (y — x — r) = o. 

4. p (p + 2 x) (p — jy 2 ) = o. 

( y — c) (y -f x 2 — c) (xy -\- cy + 1 ) = o. 

5. y+jl i =p(x+i). y + c 2 = c(x + 1). 

6. p 2 y — y + 2/jc = 0. jy 2 = 2 ^ # + <r 2 . 

EQUATIONS OF THE N™ ORDER. 
230. Special Forms. 

1. Form, 2 = /0r> 

Rule : Integrate n times with respect to x. Cf . §211. 



Differential Equations 365 

Rule : Multiply both members by 2 dy and integrate. 



Thus, let -j-*=y\ 



d 2 y 
dx 2 



then, — / dyd 2 y = 2 j ydy. 



Hence, -^ = y 2 + c. 



dy 2 
dx 2 



Hence, dx = ± 



Therefore, x = ± log ( j> + Vy 2 + <r) + /. 

Rule : ./W — =. p ; then (1) becomes 
dx 7 



/( -x^=i ' v^=a ' : * •/, *)=o 



which is of an order one lower than (1). 
To illustrate, let 

<y^ 2 / \dx) ' 

dy 
Letting — = p, and extracting square root of both members, 

we have 

$ = \la 2 - Pp 2 ; 
dx 

1 . ,bp- 

.', x = -sin l (- c. 

b a 

Hence, p = — sin (bx — c). 



366 Integral Calculus 

Integrating again, we find 

IP-y -f a cos (bx — c) = c d 
to be the general solution. 

+ --'<&'&> ■■■%»)- 

dy 
Rule : Put -— = p and change independent variable from x to y. 

Then (2) becomes 

which reduces the order 0/(2). 

_ . d 2 y d dy dy dp 

Then since — - = — —.— =— .p, 
dxr dy dx dx dy 

we have dp 2 2 

Integrating, we find 

/2^> 



Hence, ^r = 



V^y — d 2 



Integrating again, and reducing to an exponential form, we 

have 

e cx = c 1 (y+ \fc 2 y 2 - a 2 ) 

for the general solution. 



Differential Equations 367 



EXAMPLES. 

d s y 

dx z 



(l y 

1. -j^- 3 = 5 bx 2 . 12 y — bx 5 -f 6 ex 2 + c x x + <r 2 . 



2. </y + a 2 ydx 2 = o. 7 = <rsin (dra: + *i). 

3. d*y— a 2 dx 2 = A/y 2 . ^ = sec { ^ (# + c) } c x . 

d 2 y (dy\ 2 

*-d + a {i) = °- " = «* + * 

5. yd 2 y + dy 2 = dx 2 . y 2 = x 2 -\- ex + c x . 



2 /d 2 y\ 2 , /dy\ 2 2 y - , j 



6 - a i^; = i + ^- t-** 4 - 



*,<?" 



= v — Tr 1 -' (x + cf +{y+ # - aK 

dx 2 

8 - *£-($)'-**** +***-*"-"* 



368 Integral Calculus 



CHAPTER X. 

MECHANICAL APPLICATIONS. 

RECTILINEAR MOTION. 

231. Formulae. Let v = velocity, a = acceleration, and 
s = distance traversed ; then 



v — — 
dt 



a = — 
dt 



.-. s = / vdt, and / = / — (i) 

.*. v = j adt, and t = / — (2) 



ds 
It 

dv 

ds 



d 

.. dv dt d 2 s 

Also a = — - = — — = -—^ .'. s = 
dt t J t dt* 



fjadt* (3) 



232. The acceleration of a body^s velocity is constant ; find the 
velocity of the body and distance traversed in any time t. 

d 2 s 
From (3) we have — = a ; (a 

ds _ . . 

.-. by integration, — = at + C {a ) 

i.e., v = at + C. 

Suppose v = v Q when / = o ; then C = v . 

Hence^ v = v + ^/, (/^) 

is the required velocity expressed in terms of the initial velocity 
Vq, the acceleration a and time t. 



Mechanical App 369 

Integrating (af), remembering that C — v , we have 

s = i at 2 + v t + C v 
Let s = s when / = o ; then C x = s . 
Hence, s = ^ at 2 -\- v 1 -\- s , (c) 

is the required expression for the distance traversed. 

Cor. i. If we suppose the body to move from rest then 
z/ =o and s = o, 

hence, (p) and (c) become 

v = at (d) 

s = ±at 2 (e) 

Eliminating / between (d) and (e) we have 

v = V2 as (/") 

for the velocity acquired by the body in moving through the 
distance s. 

233. Falling Bodies. We know from mechanics that the 
acceleration of the velocity of a body caused by the earth's 
attraction (force of gravity) is sensibly constant. Denoting 
this acceleration by g (= 32.2 ft. a second, nearly), and the 
distance fallen through by h, we have from (a), (p), (c), § 232. 

d 2 s 
df 2= ^ 

v = v + gt (a) 

h = v t+±gt 2 + h (b) 

in which v = velocity, and h = distance traversed at the be- 
ginning of the epoch. 



370 Integral Calculus 

Cor. If the body starts from rest then v = o, and h = o ; 
hence, 

7) = erf 

Eliminating t, v = \l2gk (c) 

234. Bodies projected vertically. If a body is projected ver- 
tically downwards ; then (a), (b) § 233, give 

v = v o + gt (a) 

h = v t + ±gt* (since ^ = o) (b) 

If projected vertically upward, then since £■ = — g, a retarda- 
tion, we have 

v = v — gt (c) 

h = vj-\gt* {d) 

Cor. If v = o in (V) we have 

for the time it takes a body to rise to its highest point when 
projected upwards with a velocity v . 

235. Body Falling in a Resisting Medium. Let us consider 
the case of a body falling in the air. It has been found by 
observation that the retarding effect of the air varies with the 
square of the velocity of the body ; hence the acceleration due 
to gravity is at any instant less than g by an amount propor- 
tionate to v 2 . We may therefore write 

a=—=g— mif, 

in which m is to be determined by observation. For con- 

g 
venience let m = -=; then 



Mechanical Applications 371 



dv 


g 2 In 2 — v' 


dt * 


n 2 " * \ n 2 


.• 


g , . dv 
- *dt= • 
n l nr — v l 



Integrating and suppose v = o when / = o, we have 



or, 



g i n + v 

— t = — log 

nr zn n — v 

2 -2-t n-\-v 



n — v 



gt gt 

,n - n 



e — e 
hence, v = n (i) 

' at at v / 



n I „ n 



+ e 

An expression which gives the velocity at the end of any 

time /. 

ds 
Replacing v by its value — , § 231, and multiplying through 

by dt, we have 



ds = 



n(e n — e n ) dt 

gt gt_ 

<r + e~ " 

g 
If s = o when / = o ; then C = — log 2 = log 1. Hence, 

?< gt 

* = ?| log ^ — I (2) 

Equa. (2) enables us to determine the distance traversed in a 
given time (/). 



372 Integral Calculus 

236. Body projected into a Resisting Medium. If we sup- 
pose the body is acted upon by no force other than the resist- 
ance of the medium into which it is projected, we have from 
the preceding article, 

dv 
a = — = — wr, 
at 

in which m is to be determined by experiments made in the 
particular medium selected. From the above expression we 

have 

i dv 

dt = -; 

m ir 

hence, / = (- C. 

mv 



Let v = 7> when / = o ; then C = — 



mv, 



i /i i 

m \v v 



v 



i 

mt -j 



an expression for the velocity at the end of any time /. Re- 

ds 
placing v by its value — , and multiplying through by dt, we 

have 

dt 

ds = ; 

i 

mt -\ 

v n 



ii 

Let s = o when / = o ; then C 1 = log - 



Hence, s = — log \ v mt + i } 



m 



which gives a relation between the distance (s) and time (/). 



Mechanical Applications 373 

237. Body Falling when Gravity is variable. 

Let g = acceleration of a body at the earth surface, and let 
r = radius of the earth. Let s (> r) = distance of a body from 
the earth center, and let a = its acceleration at that instant ; 
then, according to Newton's law, viz., that the acceleration of a 
body at different distances from the earth center varies inversely 
as the square of its distance, we have 

a : g : : r 2 : s 2 . 

d*s gr 2 

Hence, * = _=-_. 

Multiplying through by ds, and integrating, we have 

ds 2 2 gr 2 _ 

Hence, — ■ = f- C, 

dr s 

2 pr 2 
i.e., v 2 = tlL- + C. 



Let v = o when s = s ; then C = 



s 

2 gr 2 



3 o 



.-. v 2 = 2 gr 2 ) S (i) 



An expression for the velocity acquired by the body in fall- 
ing from the height s . 

ds 
Replacing v by its value — , and solving for dt, we have 

_, s \i sds 



2 
6 ' 



2gr 



n/v 



Hence, t =. (-^-X \(s s- M - -° vers" 1 —l+.Q 
\2gr 2 ) ( K 2 s ) 

48 1374 



374 



Integral Calculus 



Let s = s when / = o ; then 



C = 



Hence, 



s \*s t 



2js r r 



.— 1 



vers~ x 2 = 



'0 



2 srr' 



2 TTS. 



»0 



J (j s - s*)i - -Vers- 1 — + — ° 
2 £ r *J ( 2 s o 2 



(2) 



which gives the time t for a body to fall from height s to 
height s. 

Cor. i. If in (i) we make ^ = oo and s= r, we have 

v = ^2gr 

for the velocity with which a body would strike the earth if it 

fell from an infinite distance in a vacuum. 

Since g = 32 ft., nearly, and r= 20,900,000 ft., nearly, we 

find 

v = j miles per second, nearly. 

Cor. 2. If in (2) we make s = r, we have the time for a 
body falling from the height ^ to the earth. 



CURVILINEAR MOTION. 

238. Velocity of a body down a curve in a vertical plane. 

Let ST be any curve in the plane VOX, referred to OY and 
s OX 2iS axes, OF being posi- 

tive downwards. Let P be 
the position of the body at 
any instant and let PA = 
(g) represent the acceleration 
due to gravity. Draw AC 
_l_ to the tangent PB, and 
Fig. 61. let PAC = ; then 




PC = g sin = acceleration in direction of motion. 



Mechanical Applications 375 

But if we let PB = ds, then PA = dy ; hence — = sin 0. 

ds 



Hence, §82, 



d 2 s dy . 

a¥ =g ~ds' 



dsd 2 s . 

•'• ~w = gdy - 

Integrating between limits y and y r we have 

ds* 



=g(j-y) (o 



.-. § 17, ^ 2 = 2g(y — /)- 

Comparing the last equation with (V) § 233, Cor., we see 
that the velocity acquired by a body in rolling down a curve is 
the same as it would acquire in falling freely through the vertical 
height. 

Cor. From (1) we have 

ds ds dy 



dt = 



V2 g (y — y') dy ^2g(y — _/ N 

fy ds dy 

.'. t= I -j- • . — . (2) 

J tf dy V2 g{y - /) 

is the time it takes the body to fall through the height, y — y' . 

239. Time of descent down an inverted cycloid. 
From § 238 (2), we have, 

ds dy 



-I 



dy ^2g(y-y') 



(a) 



We are to find what this expression becomes when applied 
to the cycloid. 



376 



Integral Calculus 



From the equation of the cycloid, x = a vers -1 - — V2 ay — y? f 
we obtain 

y 



dx 



dy V2 ay — jf 



But § 18, (3), 



Hence 



ds I 



dx 2 
df 



+ 1. 




This value in (a) gives t = y - j 



dy 



Let y — y = z ; then dy = dz, and 2a— y = 2 a — y r — z. 

dz 



Hence 



i.e., 



t = 




gJ \li2a- y')z- z* 



t = \/ -vers 

g 2a-y 



- 2Z -, + c. 



If we suppose the body to fall from C to B we have z = o at 
C, and z = 2a — y at j5. Hence between these limits of 2 
we have 



/ = 



.\<tm^ 



a\ , 2 a— y 
I vers x 2 



£" 



2 # — y 



7 + C — vers _1 o — C > ; 



Mechanical Applications 



377 



is the time it takes the body to fall from the position C to the 
lowest point B of the curve. Since y' is any value of y, the 
point C is any point of the cycloid ; hence the time required for a 
body to fall from any point of an inverted cycloid to its lowest 
point is constant. Theoretically, therefore, the cycloidal arc is 
the path of a pendulum which vibrates in equal times. 

240. A projectile is thrown obliquely upward with a velocity 
v ; find (i), the equation of its path ; (2), the coordinates of its 
highest point ; (3), the angle of projection in order that its range 
may be a maximum. 




Fig. 63. 

(1). Let the origin of coordinates O be the point of propul- 
sion ; DOX = angle of projection and OD = v = velocity of 
projection. Draw DE _L to OX; then 

OE = v cos = velocity in direction of X, 
DE = v sin 6 = velocity in direction of Y. 

Since no retarding force acts in the direction of X(the resist- 
ance of the air being neglected) the velocity in that direction is 
uniform. Denoting the distance traversed in that direction 
in any time / by x we have, 

x = v cos 6 1 {a) 

Denoting the distance traversed in the vertical direction in 
the same time thy y we have, § 234 (d), 

y= vsm6t-±gt\ (b) 



378 Integral Calculus 

Eliminating / between (a) and (b) we have, 

y = tan 6.x f— -j (i) 

2 zr cos' 2 y 

which expresses the relation between x and y for all values of t ; 
hence it is the equation of the trajectory. This curve is 
obviously a parabola. (Ana. Geom. p. 178.) 

(2). At the highest point C, the tangent, is || to X; hence 
dy 



= o, 



V 2 

i.e., x = — sin 2 — OA U) 

2 g 

is the abscissa of the highest point. This value of x in (1) gives 

j^J^sin 2 6 = AC 
*g 

for the ordinate of the highest point. 

(3) Denoting the range by R we have, since the curve is 
symmetrical with respect to AC, 

R = 2 OA = OB\ 



dx 




From (1), 


d y _ tan 


dx 1? cos 2 6 


• 


i' 2 
,\ x = —sin 6 cos 0, 



nee (V) 


R = — sin 2 0. 
g 


Hence, 


dR 2 7? 

— - = — cos 2 
dB g 




.'. COS 2 $ = O, 




.-. e = 45°. 



Mechanical Applications 379 

Since f"(K) is negative for = 45 ° this value of 6 corre- 
sponds to a maximum value of R. 

Cor. Since 

2 2 2 

R = - sin 2 $ = - sin (180 - 2 6) = - sin 2 (90 - 0) 

we see that the same range (R) may be attained with a 
given initial velocity (7/) under two angles of projection and 
90 — 6. We see also that these angles are complementary. 

CENTER OF GRAVITY. 

241. Definition. The center of gravity of a body is that point 
through which the line of action of the body's weight always passes. 

242. Formulae. 

Jxdv 



fydv 

y\ = 



z, = 



I 



V 

zdv 



v 

These formulae enable us to determine the coordinates 
(pc v y\,z^) of the center of gravity of any given homogeneous 
body, of volume v. 

If the body is symmetrical with reference to a plane, this 
plane may be taken as the XY plane ; whence z 1 = o. 

If the body is symmetrical with reference to a straight line, 
this line may be taken as the X-axis ; whence y x = o and 
z x = o. 

243. To find the center of gravity of a circular arc. 

Let ABC be the arc and let OX be the axis of symmetry. 



380 



Integral Calculus 



Let (x,y) [= OZ>, DA\ be the coordinates of the extremity (A) 
of the arc. Since 



we have, § 242, x x = 



dv = ds = \nix*~+~df 

Xx ydx 2 -f- </y 2 



J> (■+(!)> 



From the equation of the circle, x 2 + y 2 = a 2 , we have 




Jv-f — 



dx y 




dy x ' 




f-ASf^ 


_2ya 



Since 27 = chord AC, we see that 
the center of gravity of a circular arc 
is on its radius of symmetry and at a 
stance from its center equal to the 
fourth proportional between the arc, radius and chord. 

244. To find the center of gravity of a circular sector. 
Here dv = d 2 A = dxdy ; hence, 



OC-% — — 



rx 



xdxdy 



A 



£1 



V^2 — ^2 



xdxdy 



—■VJ _ .3.2 



Mechanical Applications 
2 I V# 2 — x 2 xdx 

= _ 



381 



A 



2 /2 3 



rd" 



If the sector is a semicircle then A = — > and 

2 



,%^-j — — 



4a 
3* 



245. 71? find the center of gravity of the area bounded by a 
parabola, its axis and one of its ordinates. 











c 




V 


B 




x' 




A 



Fig. 65. 

Let y 2 = 2px be the equation of the parabola ; then 



#! = 



xdxdy I I xdxdy 

Jo Jo 



A 



\l2fi \ x?dx 1 — 6 

Jo 2 \}2pX* 



A 



5^ 



and 



y± = 



ydxdy I I ydxdy 

Jo Jo 



£ 



A 



p I xdx 



px 2 

2~A 



382 Integral Calculus 

But, §215, 

dxdy =jj dxdy = \l2fi j x*dx 

Jo Jo Jo 

2 V 2 px% 

Hence, x x = § x and y x = §jy. 

246. To find the center of gravity of a parabolic spandrel. 

That is, to find the center of gravity of OB C, Fig. 65. 
Here, 

- 

££ xdydx _ £t xdydx _ wi" yidy / 



AAA 4op 2 A 

v 2 

ydydx I I ydydx — / yrdy 

_ Jo Jo 2 pjp _^ y* 

yi ~ ~~A ~~ ~^A ~~ ~A ~ZpA 

But A = OBC= OABC-OAB; 

i.e., A = xy — \xy = \xy. 

Hence, x x = t 3 q x, and y x = \y. 

It will be observed from the limits of integration that the area 
OR C is supposed to be generated by a line II to X moving in 
the direction of Y, the line being limited by the K-axis and 
the curve. The student may derive the same result by pro- 
ceeding as in previous articles. 

247. To find the center of gravity of a Pyramid or Co?ie. 
From § 210 (p), we have 

dV= Adx; 

1 xdV j Axdx 

hence, x x — — — = —p 

/ Adx 



Mechanical Applications 383 

Adopting the figure and notation of Ex. i, p. 334, we have 

A _ x 2 m 
~A f= W 



.:A-A /; 



hence x-. = ^-r = - h. 



£ 



x z dx 



248. To find the center of gravity of a paraboloid of revolution. 

Let y 2 = 2px be the equation of the generating curve, and let 
x = h and x = h! be the equation of two planes || to YZ. We 
wish to find the center of gravity of that portion of the para- 
boloid included between the planes. Since X is an axis of 
symmetry we have 

/ xdV 7r / xy 2 dx 

*i = --jr- = fh ' 

7r I y 2 dx 

since dF= 7ry 2 dx, § 209 (a) ; 
hence, x-, = 



2p \ xdx 



h „ VI __ l,'1 



3 h 2 - h' 
z F / xux 

Jh' 

If h' = o, then 



x \ — 3 "■■> 

i.e., The center of gravity of a segment of a paraboloid of 
revolution estimated from its vertex is two-thirds of its altitude. 

249. To find the center of gravity of the semi-ellipsoid of 

revolution. 

b 2 
Let v 2 = -5 (2 ax — x 2 ) 

be the equation of the generating curve, then 



384 



OQ-% — 



Integral Calculus 
J (2 ax — x 2 ) xdx 



J (2 ax — x 2 ) dx 



= h 



MOMENTS OF INERTIA. 

250. Definition. The moment of inertia of any area about 
an axis is the integral of the product arising by multiplying the 
differential of the area by the square of its distance from the axis. 

251. Formula. „ 

M= I t*dA 

in which M = moment of inertia, A = area, and r = distance 
of dA from the assumed axis. 

252. To fitid the moment of inertia of a rectafigle. 

Y 

d 



D 



Fig. 66. 

Let AB = b and AD = d ; then 
(1), OY being the axis. 



M 



r 2 dA 



d b 

= I I x^dxdy 

2 



& 

2 bx^ 

* 3 

2 



bd* 
12 



Ad" 



12 



Mechanical Applications 
(2), OX being the axis. 

M= J\J\fdydx 

= djjfdy= - = — • 

%j 2 

(3), OZ being the axis. 

I \ H* 2 +f)dxdy 
^ + db* A (d 2 + ^) 



385 



12 



12 



253. Moments of inertia of hollow-girders, channel-bars, and 

I-iron. 

b b ■ b 









— 


b' 

___d 


— 









b' 



4*' 



4*' 



Fig. 67. 

Let the X-axis, passing through the center of gravity of the 
section, be taken as the axis of moments ; then, in all three 
cases, we readily deduce 

bd 3 - b'd' z 



M= 



12 



254. Moment of inertia of a circle about a diameter. 
In this case, a representing the radius of the circle, 



M= f r 2 dA 

%J — a <J — 



V«2 Zl x 2 



V^^ 



x 2 dxdy 



= 2 I x 2 (a 2 — x 2 )i dx. 

U — a 



38G 



Integral Calculus 



Using the reduction formula of §§ 185, 186, we find 

7J77 4 Aa 1 

M= — == 

4 4 



Y 

1 ° 






B 

\a 


, 






C 



Fig. 68. 

255. Momejit of inertia of an ellipse about its minor-axis. 
Here M= f r 2 dA 

b V~ o 

C a Ca * " X 

= J I x^dxdy 



^ 



■rrba 8 A a 2 



DEFLECTION AND SLOPE OF BEAMS. 
256. Formula. From mechanics we have 

ir EI 
M= — > 



tt 



for the relation between the moment of the extraneous forces 



Mechanical Applications 387 

(EI\ 
{AT) and the moment of the internal resistance — about 

the neutral axis of any section. In this formula E = coeffi- 
cient of elasticity of the material of which the beam is made ; 
J = moment of inertia of section, and p — radius of curvature 
of the curve of mean fiber, at the point in which it pierces the 
section. But § 134, 

C fdy\ 2 ) % 



d 2 y 
dx 2 



EI 



Hence, 



M= 



d 2 y 
dx? 



■ + @)'i 



/dy\ 2 
Since ( — - J = tan 2 a, i.e., the square of the slope of the beam, 

and since in practice the value is small, we may omit it and 

write 

d 2 y 



M=EI 



dx'' 



(») 



Formula (2) is sufficiently accurate for all practical purposes, 
and is in general use. 

257. Slope and defleciio7i of a beam loaded at one e?id and fixed 
at the other. 

Let / = length of beam, and 
W = weight applied at its end 
A. Let OA be the mean fiber, 
and S a plane _L to OA at a 
distance x from O ; then 

M= WQ-x). 




Fig. 69. 



388 



Integral Calculus 



d 2 y W 
Hence, § 256 (2), — = ^(/- x) 



dy 

dx 



W 
2EI 



(2 Ix — x 2 ) -f- C. 



dy 



When x = o, — = o, since the tangent at O is coincident 



with X; hence C = o. 



Integrating, 



dy 
dx 

y = 



2^£Y 



(2 /jc — .x 2 ). 
(3Z* 2 -^ 3 ), 



(*) 



since when x = o, j = o and therefore C" = o. Equa. («) is 
the equation of the curve OA, i.e., the equation of the curve 
which the mean fiber takes under the action of the load W. 

If in (a) we make x = I and let 8 = value of y when x = I, 
we have 

for the maximum deflection of the beam. 

258. Shape a?id deflection of a beam fixed at one e7id and 
uniformly loaded. 




Let w = load per unit of length of beam ; then at any 
section 6" 

M= w{l- x) ^— ^ = - (/- x)\ 



Mechanical Applications 389 



d 2 y w 

Hence ' M = 7Wl ( - p - 2lx + ^ 



dy 
Since -r- = o when x = o ; .*. C = o. 
ax 

Integrating again we have 

w /Px 2 lx z x*\ 

Since y = o when x = o ; .*. C = o. Equa. (<z) gives the 
shape the beam assumes under the action of the load. Repre- 
senting the maximum deflection by 8' which obviously occurs 

when x = I we have 

a// 4 
8 = &EI 

Cor. Let W = wl = load on beam ; then 



8Ef 

Comparing 8' with 8 of § 257 we find 

8 = 8 8' = 3 8', nearly. 

That is, the deflection is nearly three times as great when the 
load is concentrated at the end as it would be if uniformly dis- 
tributed over the beam. 

259. Shape and deflection of a beam supported at both ends and 

loaded in center. 

W 
In this case M = — x\ 

2 

d 2 y W 






dx 2 2 El 
dx \EI 



390 



Integral Calculus 



If x = - , -j- = o, since at the middle of the beam the tan- 

2 dx 



gent is || to X; .*. 



C= - 



1 6 EI 




Hence, 



dx 4. EI 



W /x s Px\ 
Integrating again, y = — ~ f - - — 1 

Since x = o, y = o ; .\ C = o. 

/ J^7 3 

When x = — we have 8 = .. _ . 
2 48 is/ 



260. Shape a?id deflectio?i of a beam supported at both ends and 
imifornily loaded. 




Fig. 72. 



In this case we have 



_ x wl w , 

M = mx • ^ = — (ar — £n. 

22 2 



Hence 



7£/ 



tf^ 2 2 .£/ 



I fc'V * t i,V I « 



Mechanical Applications 



391 



hence 



w fx z /x 2 \ _ 



dy 

dx 2 JE I \$ 2 I 



wu I dy i wl z 

When x = - » — = o ; .-. C = =— 

2 ^C 2\±L1 



Hence 



</>' «/ ( x s /x 2 l z ) 

^r 2 j5"/ (3 2 12 ) 

w { x 4 lx z l z x 

• y = — ^r \ t ■• — 

2 i,/ 12 6 12 



Since # — o, _>' = o; .-. C = o. 

K* = -, then y = _|- = ^ . 

Cor. Comparing the value of 8 of § 259 with S' of this 

article, we find 

8 = fS', 

i.e., the deflection produced by a load concentrated at the cen- 
ter of a beam is f of that produced by the same load when 
uniformly distributed. 

261. Shape and deflection of a beam fixed at both ends and 
uniformly loaded. 




Fig. 73- 

This case is similar to that of the preceding except that an 
unknown moment m acts on the portion of the beam OB ; hence, 



w 



M=— (x 2 — Ix) + m. 



(0 



392 Integral Calculus 

d 2 y i ( wx 2 wlx ) , . 

Hence 7^ = ^ \- m \ • (a) 

dx 2 EI (22 ) w 

Integrating and noting that when x = o, — = o, and there- 
fore C = o, we have, 

dy 1 ( mx 3 «Vjc 2 



4 + OT * 1 ' {b) 

At the point C where x = /,—-= o. If we substitute these 

dx 

values in the last expression, we find after reduction 

wl 2 

m = ? (c) 

12 

for the moment of the unknown couple acting at the points of 
support. Substituting this value of 771 in (Jj) and integrating, 
we find, 

I ( WX* - ^/X S 7£>/ 2 X 2 ) , , 

y = ITr] + W 

EI (24 12 24 ) 

since x = o gives y = o, and therefore C f = o. 
Making jc = - in (d) we find 

, 7£// 4 ^7 3 

o — 



384^/ 384^/ 

Comparing this value of 8 with that of S' in § 260, we find 

8' = 5 S, 

that is, by fastening the ends of a beam at its points of support, 
the deflection caused by a uniform load is only one-fifth of 

what it would be if the beam merely rested on its supports. 

wl 2 . z 

Again, making m = in (1), we have 

wx 2 wlx wl 2 
M= 1 

2 2 12 



Mechanical Applications 



393 



Making x = - in this value we find, 



M= - 



wP 
24 



wV' 



Comparing this value with the value of m = we see that 

the bending moment at the point of support is twice that at the 
center, i.e., the beam is twice as strong at the center as it is at 

the points of support. 

d 2 y 
Making—-^ = o in (a) and giving m its value in (V), we have 

wx 2 wlx wP 

+ = o ; 

2212 

P 
.'. x 2 — Ix -\- — = o ; 
6 

M '~ ) 

.*. x = — { 1 +..-->: 



are the abscissas of the points of inflexion A, D, Fig. 73. 



262. To find the strongest rectangular beam that can be cut 
from a cylindrical log. 

We have from mechanics 



p= E d 
P 2 



for the stress on a unit of area at the distance 
— .from the neutral axis of a beam when under 




transverse strain. 



Hence §§ 256, 252, P = 



Md 
T2 



M 



Fig- 74- 



(a) 



in which b and d are the breadth and depth of a rectangular 
beam. It is obvious that that beam strained by a moment M 



394 Integral Calculus 

will be strongest in which P is least. But P is least when bd 2 
is greatest. Cf. (a). 

Let D be the diameter of the log ; then d 2 = D 2 — b 2 , 

.'. bd 2 = bD 2 - b s ; 

d (bd 2 ) 
Differentiating, we have, = D 2 — 3 b 2 = o, 



hence b = D y - , and d = D 



3 y 3 

are the dimensions of the strongest rectangular beam. 



ADVERTISEMENTS. 



ANALYTIC GEOMETRY 

PLANE AND SOLID. 
BY E. W. NICHOLS, 

Professor of Mathematics in the Virginia Military Institute. 



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Lefevre'S Number and its Algebra. Introductory to college courses in algebra. $1.25. 
Lyman's Geometry Exercises. Supplementary work for drill. Per dozen, $1.60;. 
McCurdy's Exercise Book in Algebra. A thorough drill book. 60 cts. 
Miller's Plane and Spherical Trigonometry. For colleges and technical schools. $1.15. 

With six-place tables, $1.40. 
Nichol's Analytic Geometry. A treatise for college courses. $1.25. 
Osborne's Differential and Integral Calculus. $2.00. 

Peterson and Baldwin's Problems in Algebra. For texts and reviews. 30 cts. 
Robbins's Surveying and Navigation. A brief and practical treatise. 50 cts. 
Schwatt's Geometrical Treatment of Curves. $1.00. 
Waldo's Descriptive Geometry. A large number of problems systematically arranged and 

with suggestions. 80 cts. 
Wells's Academic Arithmetic. With or without answers. $1.00. 
Wells's Essentials of Algebra. For secondary schools. $1.10. 
Wells's Academic Algebra. With or without answers. $1.08. 
Wells's New Higher Algebra. For schools and colleges. $1.32. 
Wells's Higher Algebra. $1.32. 
Wells's University Algebra. Octavo. $1.50. 

Wells's College Algebra. $1.50. Part II, beginning with quadratics. $1.32. 
Wells's Essentials of Geometry. (1899.) $1.25. Plane, 75 cts. Solid, 75 c' 3. 
Wells's Elements of Geometry. Revised. (1894.) $1.25. Plane, 75 cts.; Solid, 75 cts. 
Wells's New Plane and Spherical Trigonometry. For colleges and technical schools. 

$1.00. With six place tables, $1.25. With Robbins's Surveying and Navigation, $1.50. 
Wells's Essentials of Trigonometry. For secondary schools. 90 cts. With tables. 

$1.08. Plane, bound separately,. 75 cts. 
Wells's New Six-Place Logarithmic Tables. 60 cts. 
Wells's Four-Place Tables. 25 cts. 

For Arithmetics see our list of books in Elementary Mathematics. 

D.C. HEATH & CO.. Publishers, Boston, New York, Chicago 



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